MEAN

Measure of Central tendency

Posted on January 18, 2024January 18, 2024 by MEAN

Statistics is the study of data. Data is collected resources that is translated into a meaningful information. Data is a measured values and it can be classified into four different perspectives. तथ्याङक शाश्त्र भनेको डाटाको अध्ययन गर्ने गणितको एउटा खण्ड हो। डाटा भन्नाले संकलन गरिएको कच्चा संसाधन हो जसलाई अर्थपूर्ण सुचनाको रुपमा मा प्रशोधन गर्नु पर्ने हुन्छ । समग्रमा, डाटा भन्नाले मापन गरिएको मान हो र जसलाई चार फरक दृष्टिकोणका आधारमा वर्गीकृत गर्न सकिन्छ।

Based on collection (Primary and Secondary) संग्रहमा आधारित (प्राथमिक र माध्यमिक)
Based on Scale/ Measurement (Nominal, Ordinal, Interval, Ratio) मापनमा आधारित (नाम बुझाउने, क्रम बुझाउने, अन्तराल बुझाउने, अनुपात बुझाउने)
Based on nature (Qualitative and Quantitative) प्रकृतिका आधारित (गुणात्मक र मात्रात्मक)
Based on distribution (Individual, Discrete, Continuous) बर्गिकरण \ वितरणका आधारमा (व्यक्तिगत, खण्डित, निरन्तर)

Based on these data, there are two common types of statistics.

Descriptive statistics
Inferential statistics

Descriptive statistics

A statistics that collects, organize and summarize the information is called Descriptive statistics. For example bar graph and mean.

Inferential statistics

A statistics that utilize current data and predicts it for future reference, is called inferential statistics. For example hypothesis test or regression analysis.

Measure of Central tendency

Measure of central tendency लाई केन्द्रीय प्रवृत्तिको मापन भनिन्छ । यसले तथ्याङकको केन्द्रमा हुने प्रवृत्तिको एकल मान (डाटा सेटको प्रतिनिधि मान) लाई जनाउदछ जसले डाटाको सम्पूर्ण मात्रात्मक सेटको प्रतिनिधित्व गर्दछ। यस केन्द्रीय प्रवृत्तिको मापनलाई स्थान वा स्थितिको मापन पनि भनिन्छ, यसैलाई औसत मापन पनि भनिन्छ।
The Measure of central tendency is a statistic that summarizes the entire quantitative set of data in a single value (a representative value of the data set) having a tendency to concentrate somewhere in the center of the data. Therefore, the tendency of the observations to cluster in the central part of the data is called the central tendency. It measures the central location (or position) of data set. It is also known as average.

NOTE

केन्द्रीय प्रवृत्तिको मापन जहिले पनि डाटा सेटको दायरा भित्र पर्दछ। The Measure of central tendency lies somewhere within the range of the data set
डाटा लाई फरक अर्डरमा पुनर्व्यवस्थित गर्दा केन्द्रीय प्रवृत्तिको मापनमा परिवर्तन हुदैन । The Measure of central tendency remain unchanged by a rearrangement of the data set

The most common types of such central tendencies are:

मध्ययक (Mean)
मध्यिका (Median), Quartile, Decile, Percentile
रीत (Mode)}

Mean is measure of central tendency that utilize each and every data to give a single best value. The arithmetic mean or simply mean is also knows as average, which is obtained by dividing the sum of all the observations by total number of observations (summed).
It is denoted by $\bar{X}$ and define as follows.

	Individual data	Discrete data	Continuous data
Arithmetic Mean	$\bar{X}=\frac{\sum{x}}{n}$	$\bar{X}=\frac{\sum{fx}}{n}$	$\bar{X}=\frac{\sum{fm}}{n}$
Geometric Mean	$\bar{X}= \left (\prod x \right)^{\frac{1}{n}}$	$\bar{X}=\left (\prod f x \right )^{\frac{1}{n}}$	$\bar{X}=\left( \prod f m \right )^{\frac{1}{n}}$
Harmonic Mean	$\bar{X}= \frac{n}{\sum \left( \frac{1}{x}\right )}$	$\bar{X}=\frac{n}{\sum \left( \frac{f}{x}\right )}$	$\bar{X}=\frac{n}{\sum \left( \frac{f}{m}\right )}$
Weighted Mean	$\bar{X}= \frac{\sum (w.x)}{\sum w}$	$\bar{X}= \frac{\sum (w.x)}{\sum w}$	$\bar{X}= \frac{\sum (w.x)}{\sum w}$

The common type of mean are

अंकगणितिय मध्यक (AM) Arithmetic Mean
The arithmetic mean answers the question, “if all the quantities have same value, what is the value to achieve the same total?” The answer is AM. For example, let Ram has Rs 100 and Shyam has Rs 120, then the avarage amount is AM, which is answered by
$AM =\frac{a+b}{2} =\frac{100+120}{2} =Rs 110$
In the figure below, a+b is same as AM+AM.
ज्यामितीय मध्यक (GM) Geometric Mean
The geometric mean answers the question, “if all the quantities have same value, what is the value to achieve the same product?”. The geometric mean is a useful when we expect changes in data in percentages as rate of change or ratios. It is utilised in the field of finance for the purpose of determining average growth rates, which are also known as the compounded annual growth rate. For example, let Ram deposited Rs 100 in a bank, on which 80% growth in first year and 25% growth in second year, then the average profit is GM, which is answered by
$GM =\sqrt{1.80 \times 1.25}=1.50$ , the average growth is 50%
Please note that, the situation can NOT be explained by $\frac{80+25}{2} =52.5\%$
In the figure below, a*b is same as GM*GM.
हार्मोनिक मध्यक (HM)Harmonic Mean
Harmonic Mean is used to calculate average speeds of various distances covered.For example, Let Ram traveled 100km with fuel efficiency 25KM per liter and next 100km with fuel efficiency 16KM per liter, then the average fuel efficiency is HM, which is answered by
$HM =\frac{2*25*16}{25+16}=19.51$
Please note that, the situation can NOT be explained by AM or GM
Because
The full efficiency for first 100 km is $\frac{100}{25}=4$ liter
The full efficiency for second 100 km is $\frac{100}{16}=6.25$ liter
The total fuel efficiency is
$\frac{200}{4+6.25}=19.51$
भारित मध्यक (WM)Weighted Mean}
A weighted mean is a kind of average where some data points contribute more “weight” than others. If all the weights are equal, then the weighted mean equals the arithmetic mean.

Application of Mean

The mean is calculated from all data value, so it is affected by each and every value of data set. It is applicable if the data distribution represents

Quantitative data
Closed ended
Normally distributed data

Relation between AM, GM and HM

Let a and b are two non-negative numbers, then

$GM^2=AM \times HM$
$AM \ge GM \ge HM$ Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,
$AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}$
The proof are as follows:

Now, we have
$GM^2=ab$
or $GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b}$
or $GM^2=AM \times HM$
Now, we have
$AM-GM=\frac{a+b}{2}-\sqrt{ab}$
or $AM-GM=\frac{a+b-2\sqrt{ab}}{2}$
or $AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}$
or $AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}$
or $AM\ge GM$ (1)
Similarly,
$GM-HM=\sqrt{ab}-\frac{2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}$
or $GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} )$
or $GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}$
or $GM\ge HM$ (2)
Combining (1) and (2), we get
$AM\ge GM\ge HM$

Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

Visualization of AM
By the property of radius and diameter, we get that
$AM =\frac{a+b}{2}$
Visualization of GM
By the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
$GM =\sqrt{ab}$
Visualization of HM
By using proportionality, we get
Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
$\frac{GM}{a}=\frac{b}{GM}$
or $GM= \sqrt{ab}$
Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
$HM =\frac{2ab}{a+b}$
By using proportionality, we get
Triangle DRQ and ODQ are similar with QR=GM,QD= $\sqrt{ab}$ , OD= $\frac{a-b}{2}$ , so we have
$\frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}}$
or $HM= \frac{2ab}{a+b}$

Example 1

Find the mean of the numbers 3, −7, 5, 13, −2
The sum of the numbers is
$\sum X= 3 − 7 + 5 + 13 − 2 = 12$
There are 5 numbers, so n=5.
Hence, the mean of the numbers is
$\bar{X}=\frac{\sum X}{n}=\frac{12}{5}=2.4$

Example 2

Find the mean of the wages from the following data

Wages	50	70	90	110	130	150
Number of Workers	2	4	5	6	2	1

Based on the data given above, the frequency table is prepared as below.

Wages $(X)$	Number of workers $(f)$	$f.x$
50	2	100
70	4	280
90	5	450
110	6	660
130	2	260
150	1	150
	$\sum f=n=20$	$\sum f x=1900$

Based on the formula, the mean wages is
$\bar{X}=\frac{\sum fx}{n}=\frac{1900}{20}=95$

Example 3

Find the average marks from the following data

Marks of the Students	0-20	20-40	40-60	60-80	80-100
Number of Students	20	50	55	40	15

Based on the data given above, the frequency table is prepared as below.

Marks of students $(X)$	Mid value of marks $m$	Number of students $(f)$	$f.m$
0-20	10	20	200
20-40	30	50	1500
40-60	50	55	2750
60-80	70	40	2800
80-100	90	15	1350
		$\sum f=n=180$	$\sum fm=8600$

Based on the formula, the average marks is
$\bar{X}=\frac{\sum fx}{n}=\frac{8600}{180}=47.8$

Median

Median is a measure of central tendency that utilize middle portion of the data to give a single best value. The median is the middle value of the data series when the values are placed in order of magnitude (in ascending or descending order). Therefore, Median is not affected by extreme values. It is denoted by $Md$ and define as follows.

	Individual	Discrete	Continuous
Median	$M_d=\frac{n+1}{2} \text{th item}$	$M_d=\frac{n+1}{2} \text{th item}$	$M_{d-class}=\frac{n+1}{2} \text{th item}$ with $M_d=L+\frac{\frac{N}{2}-cf}{f} \times i$

Calculating the median is also very simple. Here are the steps:

Sort the data in an ascending order.
Find the middle number of the sorted data.
If there’s an odd number of data, get the value exactly in the middle.
If there’s an even number of data, get the mean of the two middle values.

Application of Median: The median doesn’t know how far the data is. It only help to split data in two parts. It is applicable if the distribution represents

Qualitative data
Open ended or Skewed data

Example 1

Find the median of the following wages(in hundreds): $40,30,35,42,32,45,48$
Given wages (in hundreds) are
$40,30,35,42,32,45,48$
Arranging the wages in ascending order, we get

30	32	35	40	42	45	48
1st item	2nd item	3rd item	4th item	5th item	6th item	7th item

Here, the number of data are n=7, thus, based on the formula, the Median is
$M_d= \left (\frac{n+1}{2} \right )^{th}$ item
or $M_d= \left (\frac{7+1}{2} \right )^{th}$ item
or $M_d= 4^th$ item
or $M_d= 40$ hundreds

Example 2

Find the median of the wages from the following data

Wages	50	70	90	110	130	150
Number of Workers	2	4	5	6	2	1

Based on the data given above, the frequency table is prepared as below.

Wages $X$	Number of Workers $f$	Cumulative frequency $cf$
50	2	2
70	4	6
90	5	11
110	6	17
130	2	19
150	1	20

Here, the number of data are $n=20$ , thus, based on the formula, the Median is
$M_d= \left (\frac{n+1}{2} \right )^{th}$ item
or $M_d= \left (\frac{20+1}{2} \right )^{th}$ item
or $M_d= 10.5^th$ item
or $M_d= 90$

Example 3

Find the median marks from the following data

Marks of the Students	0-20	20-40	40-60	60-80	80-100
Number of Students	20	50	55	40	15

Based on the data given above, the frequency table is prepared as below.

Marks of the Students $X$	Number of Students $f$	Cumulative frequency $cf$
0-20	20	20
20-40	50	70
40-60	55	125
60-80	40	165
80-100	15	180

Here, the number of data are $n=180$ , thus, based on the formula, the Median class is
Md Class $= \left (\frac{n}{2} \right )^{th}$ item
or Md Class $= \left (\frac{180}{2} \right )^{th}$ item
or Md Class $= 90^th$ item
Here, $90^th$ item lies in the $cf$ of 125, thus
$L=40,f=55, cf=70,i=20$
Hence, the Median is
$M_d=L+\frac{\frac{N}{2}-cf}{f} \times i$
or $M_d=40+\frac{\frac{180}{2}-70}{55} \times 20=47.27$

Example 4

Find the median marks from the following data

Marks of the Students	0-20	20-40	40-60	60-80	80-100
Number of Students	2	3	5	4	6

Based on the data given above, the frequency table is prepared as below. \

Marks of the Students $X$	Number of Students $f$	Cumulative frequency $cf$
0-20	2	2
20-40	3	5
40-60	5	10
60-80	4	14
80-100	6	20

Here, the number of data are $n=20$ , thus, based on the formula, the Median class is
Md Class $= \left (\frac{n}{2} \right )^{th}$ item
or Md Class $= \left (\frac{20}{2} \right )^{th}$ item
or Md Class $= 10^th$ item
Here, $10^th$ item lies in the $cf$ of 10, thus
$L=40,f=5, cf=5,i=20$
Hence, the Median is
$M_d=L+\frac{\frac{N}{2}-cf}{f} \times i$
or $M_d=40+\frac{\frac{20}{2}-5}{5} \times 20=60$
NOTE
In the example above, student may ask that the median $60$ does not lie in the class $40-60$ as instructed for inclusive data groupings, teaches need to encourage the usual rules for computing.

Quartile, Decile and Percentile

The formula for Quartile, Decile and Percentile are similar as of Median.

	Individual	Discrete	Continuous
Quartile k=1,2,3	$Q_k=\frac{k(n+1)}{4} \text{th item}$	$Q_k=\frac{k(n+1)}{4} \text{th item}$	$Q_{k-class}=\frac{k(n)}{4} \text{th item}$ with $Q_k=L+\frac{\frac{kn}{4}-cf}{f} \times i$
Decile $k=1,2,\cdots 9$	$D_k=\frac{k(n+1)}{4} \text{th item}$	$D_k=\frac{k(n+1)}{4} \text{th item}$	$D_{k-class}=\frac{k(n)}{4} \text{th item}$ with $D_k=L+\frac{\frac{kn}{4}-cf}{f} \times i$
Percentile $k=1,2,\cdots 99$	$P_k=\frac{k(n+1)}{4} \text{th item}$	$P_k=\frac{k(n+1)}{4} \text{th item}$	$P_{k-class}=\frac{k(n)}{4} \text{th item}$ with $P_k=L+\frac{\frac{kn}{4}-cf}{f} \times i$

Mode

The concept of mode, as a measure of central tendency, is preferable when it is desired to know the most typical value, e.g., the most common size of shoes, the most common size of a ready-made garment, the most common size of income, the most common size of pocket expenditure of a college student, the most common size of a family in a locality, the most common duration of cure of viral-fever, the most popular candidate in an election, etc.
Thus, Mode is a measure of central tendency that utilize fashionable (most repeated data) information to give a single best value. So, Mode is an average value which occurs most frequently in a set of data i.e. it indicates the most frequent (common) results. It is not affected by every values. It is denoted by $Mo$ and define as follows.

	Individual	Discrete	Continuous
Mode	Repeated data	Repeated data/ Table analysis	$M_0=L+\frac{f_1-f_0}{2f_1-f_0-f_2} \times i$ $M_0=L+\frac{f_2}{f_0+f_2} \times i$

Application of Mode: The mode doesn’t know anything about any number in the collection but the one which appears most frequently. It is best applicable when concerning about

Frequency related data
Fashionable data

Example 1

Find the mode value of the following data: $3, 7, 5, 13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29$
Given data set are
$3, 7, 5, 13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29$
In frequency table, the data set becomes

X	3	5	7	12	13	14	20	23	29	39	40	56
f	1	1	1	1	1	1	1	4	1	1	1	1

Being highest frequency 4, the mode value is $23$ .

Example 2

Find the Mode of the wages from the following data\par

Wages	50	70	90	110	130	150
Number of Workers	2	4	5	6	2	1

Being highest frequency 6, the mode value is $110$ .

Example 3

Find the Mode of from the following data

Wages	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Number of Workers	4	12	15	18	32	14	13

Being highest frequency 4, the model class is $40-50$ . Thus,
$L=40,f_0=18,f_1=32,f_2=14,i=10$
Hence, using formula, the Mode is
$M_0=L+\frac{f_1-f_0}{2f_1-f_0-f_2} \times i$
or $M_0=40+\frac{32-18}{2 \times 32-18-14} \times 10=44.47$

Analytical method to find the Mode

If the frequency distribution is regular, then mode is determined by the value corresponding to maximum frequency. There may be a situation where frequency distribution is NOT regular, means the concentration of observations around a value having maximum frequency is less than the concentration of observations around some other value. In such a situation, mode cannot be determined by the use of maximum frequency criterion. Further, there may be concentration of observations around more than one value of the variable and, accordingly, the distribution is said to be bi-modal or multi-modal depending upon whether it is around two or more than two values. In such cases, we use analytical method (also called tabular or grouping or empirical method) to find the Mode.
दिएको श्रेणिमा Mode अस्पष्ट भएमा वा तलका निम्न अवस्थामा यो बिधीको प्रयोग गरिन्छ ।

highest frequency सख्या एक भन्दा बढी भएमा
highest frequency तथ्याङकको सुरु वा अन्यतिर भएमाा
highest frequency को वरिपरि ठुला frequency भएमाा
frequency को अनियमित घटबढ भएमाा

यस अवस्थामा Mode पत्ता लगाउन Empirical Method (Mode=3 Median -2 Mean) वा analytical method प्रयोग गर्न सकिन्छ । तर यि दुबै बिधीमध्ये analytical method लाई बढी बिश्वासनिय मानिन्छ।

Example 4

Find the Mode of from the following data

Wages	10	20	30	40	50	60	70	80	90
Number of Workers	1	5	17	22	21	20	9	3	4

Here, the maximum frequency is $22$ , however three are big frequencies around 22, thus we use analytical method to find the Mode.
Hence, based on the rule, the analytic table is given as below.

Wages	$f$	1st + 2nd	2nd+ 3rd	1st+2nd+3rd	2nd+3rd+4th	3rd+4th+5th
10	1
		6
20	5			23
			22
30	17				44
		39
40	22					60
			43
50	21			63
		41
60	20				50
			29
70	9					32
		12
80	3			16
			7
90	4

Prepare a table consisting of 7 column, 1st column for X, 2nd column for frequencies of X.
In third column, add the frequencies, starting from the top and grouped in twos.
In forth column, add the frequencies, starting from the second and grouped in twos.
In fifth column, add the frequencies, starting from the top and grouped in threes .
In sixth column, add frequencies, starting from the top second and grouped in threes.
In seventh column, add the frequencies, starting from the top third and grouped in threes.
Finally, prepare frequency chart based on the analytic table

Based on the analytic table, the frequency chart is prepared as below.

Column	10	20	30	40	50	60	70	80	90
1				1
2					1	1
3				1	1
4				1	1	1
5					1	1	1
6			1	1	1
Total			2	4	5	3	1

Here, the highest frequency is aligned with 50, therefore, Mode=50.

Relation between Mean Median and Mode

A distribution in which the values of mean, median and mode coincide (i.e. mean = median = mode) is known as a symmetrical distribution.
Conversely, when values of mean, median and mode are not equal the distribution is known as asymmetrical or skewed distribution. In moderately skewed or asymmetrical distribution, a very important relationship exists among these three measures of central tendency. In such distributions
Mode = 3 Median – 2 Mean

Symmetrical Distribution

Sequence and Series

Posted on January 17, 2024January 18, 2024 by MEAN

In mathematics, the word, “sequence” we usually mean a ordered collection with an identified first member, second member, third member and so on. is also called progression (AP).

Introduction

Sequence is a pattern of ordered numbers. Since the numbers follow a pattern, we can relate each number to its numerical position with a rule. Such a rule is called sequence. Each number in a sequence is a term of the sequence.

Presumably, a sequence continues by following the pattern that first few “terms” suggest. To understand the pattern more explicitely, it is also useful to think of a sequence as a function.
Thus,
So, a real-valued function defined on N = {1, 2, 3, . . .} is a sequence.
The range of the function is real numbers, the term of sequences
Mathematically, it is written as
$f : N \to R$ .

A sequence can be defined with two different ways

Recursive definition (Syntactic definition), Implicit definition
Formal definition (Semantic definition), Explicit definition

Recursive definition

A sequence can be described by comparing each term to the one that comes before it, i.e., by defining the later term in relationto previous terms. Such rule of describing sequence is called recursive definition. The recursive definition contains two parts. twwo parts, the initial condition and definition. For example, in a sequence
133,130,127,124,…,
each term after the first term is equal to three less than the previous term.
Therefore, a recursive definition for this sequence is as follows

an initial condition (the value of the first term): a₁=133
a recursive definition (relates each term after the first term to the one before it): a_n=a_n-1-3 for n>1

Example 1

Writing a recursive definition for a sequence

The number of blocks in two dimensional pyramid is a sequence that follows a recursive formula. What is the recursive definition of the sequence?

The solution is as follows.

Let us count the number of blocks in each pyramid: it is 1,3,6,10,….
Now, subtract consecutive terms to find out what happens from one term to the next
a₂-a₁=3-1=2
a ₃-a₂=6-3=3
a ₄-a₃=10-6=4
Now, use n to express the relationship between successive terms
a _n-a_n-1=n
To write the recursive definition, state the initial condition and the recursive formula
a ₁=1 and a_n=a_n-1+n

Example 2

What is the 100th term of the pyramid sequence in the example given below?

Solution
To find the explicit formula, expand the first few terms of the pyramid sequence. which is as below.

a₁	a₂	a₃	a₄	…	a_n
1	3	6	10	…	a_n
1	1+2	1+2+3	1+2+3+4	…	1+=2+3+4+…+n

Therefore,
a _n=1+2+3+4+—+(n-2)+(n-1)+n (1)
Writing in reverse order, we get
a _n=n+(n-1)+(n-2)+…+4+3+2+1 (2)
Adding (1) and (2), we get

a_n	=1	+2	3	+—+	(n-2)	+(n-1)	+n
a_n	=n	+(n-1)	+(n-2)	+—+	+3	+2	+1
2a_n	=(n+1)	+(n+1)	+(n+1)	+—+	+(n+1)	+(n+1)	+(n+1)
2a_n	=n(n+1)
a_n	= $\frac{n(n+1)}{2}$

Therefore, the explicit formula, for this sequence is
a _n= $\frac{n(n+1)}{2}$
Now, we substitute n by 100 to find the 100th term. Which is
a _n= $\frac{n(n+1)}{2}$
or a ₁₀₀= $\frac{100(100+1)}{2}$
or a ₁₀₀=5050
Now, we define sequence mathematically.

Semantic Definition

A sequence is a function defined on the set $\mathbb{N}$ . For example, f(a_n)=3a_n. There are different types of such sequences. Among them we discuss three basic types of sequence in the following section.

Arithmetic sequence
Geometric sequence
Harmonic sequence

Arithmetic Sequence

An arithmetic sequence is a list of numbers with a definite pattern based on addition. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.

In the arithemetic sequence
The constant difference in all pairs of consecutive or successive numbers is called the common difference. It is denoted by the letter d.
Please note that
Difference here means the second minus the first. We add the common difference to go from one term to another.

Therefore,
An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For example,
the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common difference of 2.

General term of arithmetic sequence

An arithmetic sequence with a starting value $a$ and common difference $d$ is a sequence of the form
$a,a+d,a+2d,a+3d,\cdots$
A recursive definition for this sequence has two parts
$t_1=a$ : initial condition
$t_n=t_{n-1}+d$ for n>1:recursive formula
Therefore, an semantic definition for this sequence is a single formula as general terms given following calulations
$t_n=a+(n-1)d$ for n > 1

Proof.

$t_1=a$
$t_2=a+(d)$
or $t_2=a+(2-1)d)$
$t_3=a+(2-1)d+d$
or $t_3=a+(3-1)d)$
$t_4=a+(3-1)d+d$
or $t_4=a+(4-1)d)$
Similarly,
$t_n=a+(n-1)d$ for n > 1

Arithmetic mean

Let a, b, and c are three terms in an arithmetic sequence, then b is called Arithemetic mean of a and c.
In this case
, first two terms a and b will have the difference which will be equal to the next two terms b and c.
So we can Thus,
$b-a = c-b$ .
Rearranging the terms, we get
or $2b = a + c$
or $b = \frac{a+c}{2}$

Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence, denoted by $S_n$
To find the explicit formula for the S_n: Sum of arithmetic sequence up to n th term, we write

$S_n$	=a	+(a+d)	+(a+2d)	+…	+[a+(n-2)d]	+[a+(n-1)d]
$S_n$	=[a+(n-1)d]	+[a+(n-2)d]	+…	+(a+2d)	+(a+d)	+a
$2S_n$	=[2a+(n-1)d]	+[2a+(n-1)d]	+…	+…	+[2a+(n-1)d]	+[2a+(n-1)d]

Thus

$2S_n=n[2a+(n-1)d])$
or $S_n=\frac{n}{2}[2a+(n-1)d]$

Example 3

If the first term of an AP is 67 and the common difference is -13, find the sum of the first 20 terms.
Here, a = 67 and d= -13, thus
S_n = $\frac{n}{2}[2a+(n-1)d]$
or S_n = $\frac{20}{2}[2 \times 67+(20-1)(-13)]=-1130$
So, the sum of first 20 terms is -1130.

Example 4

[Modeling]:Look at the figure below. The length of the side of each cube is 1cm. Copy the figure in isometric dot paper Based on the pattern

Draw Figure no. 4 of this pattern on the dot paper
Find the volume of the four Figures
What would be the volume of Figure no. 12 of this pattern
Write an equation to represent the volume of Figure n

Example 5

Show that sum of first n odd natural number is S_n= $n^2$
Here, a=1,d=2, thus the sum is
S_n= $\frac{n}{2} [2a+(n-1)d]$
or S_n= $\frac{n}{2} [2.1+(n-1)2]$
or S_n= $\frac{n}{2} [2+2n-2]$
or S_n= $n^2$

Example 6

Show that sum of first n even natural number is S_n=n(n+1)
Here, a=2,d=2, thus the sum is
S_n= $\frac{n}{2} [2a+(n-1)d]$
or S_n= $\frac{n}{2} [2.2+(n-1)2]$
or S_n= $\frac{n}{2} [4+2n-2]$
or S_n=n(n+1)

Geometric sequence

Euclid’s book The Elements (300 BC, Book VIII) introduces a “geometric progression” as a progression in which the ratio of any element to the previous element is a constant.
The Greeks, over two thousand years ago, considered sequences such as
$\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},....,\frac{1}{2^k},...$
and their sums, such as
$\displaystyle \sum_{k=1}^{k=\infty} \frac{1}{2^k}$
The sum, above, at any finite step, is always less than the number 1.
Since the sum is less than 1 at any finite step , we can conclude that the series converges in the limit to 1

Does a series have a sum?
$\sum_{n=1}^\infty \frac{1}{2^n}$
Answer
$\sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...=1$
The sum of the series is 1.
The visualization is as follows
Imagine that we paint a blank canvas in steps. At each step, we paint half of the unpainted area. The total area painted after “n” steps is therefore the “n”th partial sum
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+ \frac{1}{2^n}$
The total area remaining unpainted is
$\frac{1}{2^n}$
After an infinite number of steps we will have painted all of the canvas, of which the area is 1.

Geometric sequence

An geometric sequence is a list of numbers with a definite pattern based on multiplication. If you take any number in the sequence then divide it by the previous one, and the result is always the same or constant then it is an geometric sequence.

In the geometric sequence,
The constant ratio in all pairs of consecutive or successive numbers in a sequence is called the common ratio. It is denoted by a letter r.
Please note that
Ratio here means the second divide the first. We use the common ratio to go from one term to another.

Therefore, a geometric sequence is a sequence of numbers in which the ratio between the consecutive terms is always constant. For instance,
the sequence 5, 15, 45,135, . . . is an geometric progression with common ratio of 3.

General term of geometric sequence

A geometric sequence with a starting value a and common ratio r is a sequence of the form
$a,ar,ar^2,ar^3,...$
A recursive definition for this sequence has two parts
$t_1=a:$ initial condition
$t_n=t_{n-1}r$ for n>1 recursive formula
Therefore, an explicit definition for this sequence is a single formula as general terms given by
$t_n=ar^{n-1}$ for n > 1

Example 1

When a ball bounces as shown below, the height of consecutive bounces become a geometric sequence. If $a_1=100,a_3=49$ , thn what is the height of 5th bounce?

The height of the first and third bounces are given in the figure above.

Solution
$a_1=100,a_3=49$
Use the explicit formula to relate a₁ to a₃ and to find r.
or a_n= $a_1 r^{n-1}$
or $a_3=a_1 r^{3-1}$
or $49=100 r^2$
or $r=\frac{7}{10}$
Find the fifth using explicit formula, thus we have
$a_5=a_1 r^4$
or $a_5=100 \times \left ( \frac{7}{10} \right ) ^4 \approx 24$

Geometric mean

In the geometric sequence, if a, b and c are three consecutive terms, then b is called geometric mean of a and b.
In this sequence, the first two terms a and b will have the ratio which will be equal to the next two terms b and c.
So we can say,
$\frac{b}{a}=\frac{c}{b}$
Rearranging the terms, we get
$b^2 = a \times c$
or $b = \sqrt{ac}$

Geometric Series

A geometric series is the sum of the terms of an geometric sequence, denoted by $S_n$
Let a geometric sequence is given by
$a,ar,ar^2,ar^3,...$
If we are adding up the terms of the geometric sequence given above, then we have a geometric sum or geometric series given by
$\displaystyle \sum_{k=0}^{k=\infty} ar^k$

Method 1

The explicit formula for the sum of finite terms is given as below
Sum of geometric sequence up to n th term: $a,ar,ar^2,ar^3,...ar^{n-1}$ is written as
S_n= $a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]$
Taking a common, we get
S_n= $a[1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]$
Multiplying by $\frac{1-r}{1-r}$ , we get
S_n= $a \frac{1-r}{1-r} [1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]$
or $S_n=\frac{a(r^n-1)}{r-1}$

Method 2

Sum of geometric sequence up to n th term: $a,ar,ar^2,ar^3,...ar^{n-1}$ is written as
S_n= $a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]$
Multiplying by r, we get
rS_n= $(a r)+(a r^2)+(a r^3)+---+[a r^{n-1}]+[a r^n]$
Substracting , we get
rS_n-S_n= $ar^n-a$
or $S_n=\frac{a(r^n-1)}{r-1}$

Infinite Sum

We know that, sum of the first n terms of a geometric series with first term a and common ratio r is
$S_n=\frac{a(r^n-1)}{r-1} ; r \ne 1$
In the case when r has magnitude less than 1, the term $r^n$ approaches 0 as n becomes very large. So, in this case, the sequence of partial sums S1,S2,S3,... has a limit:
So,
$S_\infty= \displaystyle \lim_{n \to \infty} \frac{a(r^n-1)}{r-1} = \frac{a}{1-r}$
The limiting sum is usually referred to as the sum to infinity of the series and denoted by S∞. Thus, for a geometric series with common ratio r such that |r | < 1, we have
$S_\infty= \frac{a}{1-r}$

Example 2

A person saves NRs100 in a bank account at the beginning of each month. The bank offers a return of 12% compounded monthly.
(a) Determine the total amount saved after 12 months.
Solution
Let
S_n: Sum of geometric sequence up to n th term.
It is written as
S_n= $\frac{a(1-r^n)}{1-r}$
Based on the formula, the solution of above problem with
$a = 100, r=1.12, n = 12$
The solution is
$S_{12}=\frac{a(r^n-1)}{r-1}= 1268.25$

Example 3

What is the sum of the finite geometric series 3+6+12+24+...+3072
Given that, the first term is 3, the common ratio is 2, and the nth term is 3072, therefore, we use explicit formula to find the n

a_n= $a r^{n-1}$
or $3072=3 \times 2^{n-1}$
or $1024=2^{n-1}$
or $n=11$
Therefore, the sum up to 11th term is
$S_n= \frac{a(1-r^n)}{1-r}$
or $S_{11}= \frac{3(1-2^{11})}{1-2}$
or $S_{11}= 6141$

Harmonic sequence

In mathematics, a harmonic sequence formed by taking the reciprocals of an arithmetic progression.
The sequence 1,2,3,4,5,6,... is an arithmetic progression, so its reciprocals
$\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$
is a harmonic sequence.

If each term of an harmonic sequence is multiplied or divide by a constant, the sequence of the resulting number are also harmonic sequence. For example,

if $a,b,c,d, \cdots$ is harmonic sequence, then

then $ka,kb,kc,kd, \cdots$ is harmonic sequence where k is constant and $k \ne 0$
$\frac{a}{k},\frac{b}{k},\frac{c}{k},\frac{d}{k},\cdots$ is a harmonic sequence where $k \ne 0$

General term of Harmonic sequence

Let $a,a+d,a+2d,a+3d, \cdots$ are the terms of arithemetic sequence
Then
terms of Harmonic sequence are given by
$\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}+\frac{1}{a+3d},\cdots$
Now,
The nth term of the sequence is given by
$a_n=\frac{1}{a+(n-1)d}$

Harmonic Mean

Harmonic mean is finding taking the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:
Harmonic Mean = $\frac{2}{(\frac{1}{a})+(\frac{1}{b})}=\frac{2ab}{a+b}$
Where a,b are the values of arithmetic sequence.

Sum of Harmonic Series

If $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}$ is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]$
where,
“a” is the first term of A.P
“d” is the common difference of A.P
“ln” is the natural logarithm

Proof
Given $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}$ are in harmonic progression,
We set the values as follows,
a=1,d=1
Then, $\frac{1}{1},\frac{1}{1+1},\frac{1}{1+2.1},…,\frac{1}{1+(n-1)1}$ are the terms
$1,\frac{1}{2},\frac{1}{3}, ...,$ are the terms
Now, the Riemann sum of the function $f(x)=\frac{1}{x}$ approximates the sum of the harmonic series given above
Therefore,
We set the values as follows,
$S_n= \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
For any common difference,the formula becoms
$S_n=\frac{1}{d} \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
or $S_n= \frac{1}{d} \int_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
or $S_n= \frac{1}{d} \left [ \ln x \right]_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d}$
or $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]$
Therefore The formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]$

A.M, G.M, H.M and their relations

A finite sequence consisting more than two terms has one or more terms in between the first and last terms. Theses between terms are called means of the sequence. Precisely

if a,b,c,d are in arithmetic sequence then b,c are arithmetic means (AM)
So, given n terms, the AM is
$AM=\frac{a+b+ \cdots}{n}$
item if a,b,c,d are in geometric sequence then b,c are geometric means (GM)
So, given n terms, the GM is
$GM=\sqrt[n]{a \times \times \cdots }$
if a,b,c,d are in harmonic sequence then b,c are harmonic means (HM)
So, given n terms, the HM is
$HM=\frac{n}{(\frac{1}{a})+(\frac{1}{b})+ \cdots}$

Theorem 1

Given any two numbers a and b the AM, GM, and HM are as follows.

AM= $\frac{a+b}{2}$
GM= $\sqrt{ab}$
HM= $\frac{2ab}{a+b}$

The proof are as follows:

Let AM is the single mean between a and b, then
AM-a=b-AM
or 2AM=a+b
or AM= $\frac{a+b}{2}$
Let GM is the single mean between a and b, then
$\frac{GM}{a}=\frac{b}{GM}$
or $GM^2=ab$
or $GM=\sqrt{ab}$
Let HM is the single mean between a and b, then
$\frac{1}{HM}-\frac{1}{a}=\frac{1}{b}-\frac{1}{HM}$
or $\frac{2}{HM}=\frac{1}{a}+\frac{1}{b}$
or $HM=\frac{2ab}{a+b}$

Theorem 2

Let a and b are two non-negative numbers, then

$GM^2=AM \times HM$
$AM \ge GM \ge HM$ Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,
$AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}$
The proof are as follows:

Now, we have
$GM^2=ab$
or $GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b}$
or $GM^2=AM \times HM$
Now, we have
$AM-GM=\frac{a+b}{2}-\sqrt{ab}$
or $AM-GM=\frac{a+b-2\sqrt{ab}}{2}$
or $AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}$
or $AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}$
or $AM\ge GM$ (1)
Similarly,
$GM-HM=\sqrt{ab}-\frac{2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}$
or $GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} )$
or $GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}$
or $GM\ge HM$ (2)
Combining (1) and (2), we get
$AM\ge GM\ge HM$

Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

Visualization of AM
By the property of radius and diameter, we get that
$AM =\frac{a+b}{2}$
Visualization of GM
By the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
$GM =\sqrt{ab}$
Visualization of HM
By using proportionality, we get
Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
$\frac{GM}{a}=\frac{b}{GM}$
or $GM= \sqrt{ab}$
Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
$HM =\frac{2ab}{a+b}$
By using proportionality, we get
Triangle DRQ and ODQ are similar with QR=GM,QD= $\sqrt{ab}$ , OD= $\frac{a-b}{2}$ , so we have
$\frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}}$
or $HM= \frac{2ab}{a+b}$

Solved Example

Compare the sum of n terms of the series: $1+2a+3a^2+4a^3+........ .$ and $a+2a+3a+ 4a ...$ up to n term
Solution
Given that
$S_n=1+2a+3a^2+4a^3+\cdots+na^{n-1}$
$T_n=a+2a+3a+ 4a +\cdots+na$
Now, substracting, we get
$S_n-T_n= (1+2a+3a^2+4a^3+\cdots+na^{n-1})-(a+2a+3a+ 4a +\cdots+na)$
or $S_n-T_n= (1+2a+3a^2-a-2a-3a)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)$
or $S_n-T_n= (1+3a^2-a-3a)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)$
or $S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)$

Case 1: If a=1, then
$S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)$
$S_n-T_n= (0)+(0)+(0)+\cdots+(0)$
or $S_n-T_n= 0$
or $S_n =T_n$ (1)
Case 1: If a >1, then
$S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)$
$S_n-T_n= (+ve)+(+ve)+(+ve)+\cdots+(+ve)$
or $S_n-T_n= Positive$
or $S_n -T_n>0$
or $S_n >T_n= 0$ (2)

Thus, the comparision is
$S_n = T_n$ if a =1
$S_n > T_n$ if a > 1

Exercise

Integration to Investment and Growth: Determine the monthly repayments needed to repay a 100000 loan which is paid back over 25 years when the interest rate is 8% compounded annually[Ans: 780.66]
Show that if three quantities form any two of the three sequences AS,GS,and HS then they also form the remaining third sequence
The sum of three numbers in AP is 36. When the numbers are increased by 1,4,43 respectively, the resulting numbers are in GP, find the numbers.
Divide 69 in three parts in AP and the product of first two parts is 483
If a be the AM between b and c, b be the GM between c and a, then prove that c will be HM between a and b.
Show that b^2 is greater than, equal to, or less than according as a,b,c are in AP, GP or HP
If a and b are two positive numbers then prove that AM,GM,HM form GP.
If the $4^{th}$ and $15^{th}$ terms of an arithmetic series are 11 and 44 respectively then, find the sum of its first 20 terms. (Ans:610)
Puja gets an employment of Rs.25000 in a month with an annual increment of Rs. 1500. How much does she earn in six years? Find her salary at 12^{th} year. (Ans: Rs.190500, Rs.41500)
In a geometric series, if the sixth term is 16 times the second term and the sum of first seven terms is $\frac{127}{4}$ , then find the sum of the first 15 term. (Ans: $\frac{32767}{4}$ )
There are 8 bags full of books. The numbers of books in each bag form a GP. If the $4^{th}$ and $6^{th}$ bags contain 24 and 96 books respectively, find the number of the books in the $1^{th}$ and last bags. ( Ans: 3, 384)
One side fo a square is 10 cm . The mid point of its sides are joined to form another square, whose mid point are again joined t form one more square. The process is continued indefinitely. Find the sum of the areas of all the squares so formed. (Ans:200 square cm)

Exercise: Miscellaneous Exercise [NCERT, pg 147]

If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and $\displaystyle \sum_{x=1}^{n} f(x)=120$ ,find the value of n.
Solution 👉 Click Here

Given that
$f(x+y)=f(x)f(y)$
or $f(x+1)=f(x)f(1)$ Substituting y=1
or $\frac{f(x+1)}{f(x)}=f(1)$
or $\frac{f(x+1)}{f(x)}=3$
It shows that the ratio of two conecutive terms is 3, therefore, the sequence is a GP, with a=3 and r=3
Now,given that
$\displaystyle \sum_{x=1}^{n} f(x)=120$
or $f(1)+f(2)+f(3)+...+f(n)=120$
or $f(1) \left [\frac{f(1)}{f(1)}+\frac{f(2)}{f(1)}+\frac{f(3)}{f(1)}+...+\frac{f(n)}{f(1)} \right ]=120$
or $f(1)[1+3+3^2+...+3^{n-1}]=120$
or $3[1+3+3^2+...+3^{n-1}]=120$
or $[1+3+3^2+...+3^{n-1}]=40$
or $\frac{a(r^n-1)}{r-1}=40$ with a=1, r=3
or $\frac{1(3^n-1)}{3-1}=40$
or $3^n-1=80$
or $3^n=81$
or $3^n=3^4$
or $n=4$
This completes the solution.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution 👉 Click Here

Solution
Let the three numbers be $a,ar,ar^2$
So, the number forms a GP with sum 56, thus
$a+ar+ar^2=56$
or $a(1+r+r^2)=56$ (1)
According to question
$a-1,ar-7,ar^2-21$ form AP
or $2(ar-7)=a-1+ar^2-21$
or $a(1-2r+r^2)=8$ (2)
Using (1) and (2), we get
$r=2,\frac{1}{2}$
If r=2, then a=8. If $r=\frac{1}{2}$ then a=32.
So, the three numbers $a,ar,ar^2$ are either 8,16,32 or 32,16,8.
This completes the solution.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Solution 👉 Click Here

Solution
Let theterms of GP are $a_1,a_2,a_3,...,a_{2n}$
According to question
$a_1+a_2+a_3+...+a_{2n}=5(a_1+a_3+...+a_{2n-1})$
or $\frac{a[r^{2n}-1]}{r-1}=5\frac{a[(r^2)^n-1]}{1-r^2}$
or $\frac{1}{1}=5.\frac{1}{1+r}$
or $1+r=5$
or $r=4$
This completes the solution.
If $\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$ (x≠ 0), then show that a, b, c and d are in G.P.
Solution 👉 Click Here

Solution
Given that
$\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
Using componendo dividendo rule, we get
$\frac{2a}{2bx}=\frac{2b}{2cx}=\frac{2c}{2dx}$
or $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
This shows that a,b,c, d are in GP This completes the solution.
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^2R^n = S^n$ .
Solution 👉 Click Here

Solution
Let the n terms of a GP are $a,ar,ar^2,ar^3,...,ar^{n-1}$
Then
$S= a+ar+ar^2+ar^3+...+ar^{n-1}$
or $S= \frac{a(r^n-1)}{r-1}$
Also,
$P= a.ar.ar^2.ar^3.....ar^{n-1}$
or $P= a^nr^{\frac{n(n-1)}{2}}$

And,
$R= \frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+\frac{1}{ar^3}+...+\frac{1}{ar^{n-1}}$

or $R= \frac{1}{a}\left [1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+...+\frac{1}{r^{n-1}} \right ]$

or $R= \frac{1}{a}\left [ \frac{1. (\frac{1}{r}^n-1)}{\frac{1}{r}-1} \right ]$

or $R=\frac{1-r^n}{(1-r)ar^{n-1}}$
Now

$\frac{S}{R}= \frac{\frac{a(r^n-1)}{r-1} }{\frac{1-r^n}{(1-r)ar^{n-1}}}$

or $\frac{S}{R}=a^2r^{n-1}$

or $\left (\frac{S}{R}\right )^n=a^{2n}r^{n(n-1)}$

or $\left (\frac{S}{R}\right )^n= \left (a^n r^{\frac{n(n-1)}{2} } \right )^2$

or $\left (\frac{S}{R}\right )^n= P^2$
This completes the solution.
If a, b, c, d are in GP, prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in GP
Solution 👉 Click Here

Solution
Given that a, b, c, d are in GP
$a=a, b=ar,c=ar^2,d=ar^3$
We have to show that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in GP
Thus we show
$(b^n + c^n)^2=(a^n + b^n).(c^n + d^n)$
or $\left [(ar)^n + (ar^2)^n \right ]^2=\left [a^n + (ar)^n\right ].\left [(ar^2)^n + (ar^3)^n\right ]$
or $[(ar)^n]^2 \left [1+ r^n \right ]^2= a^n\left [1 + r^n\right ].(ar^2)^n \left [1 + r^n\right ]$
or $a^{2n}r^{2n} \left [1+ r^n \right ]^2= a^{2n}r^{2n} \left [1 + r^n\right ].\left [1 + r^n\right ]$
or $a^{2n}r^{2n} \left [1+ r^n \right ]^2= a^{2n}r^{2n} \left [1 + r^n\right ]^2$
It shows that, $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in GP
This completes the solution.
If a and b are the roots of $x^2 - 3x + p = 0$ and c, d are roots of $x^2 - 12x + q = 0$ , where a, b, c, d form a GP. Prove that (q + p) : (q – p) = 17:15.
Solution 👉 Click Here

Solution
Given that a, b, c, d are in GP
$a=a, b=ar,c=ar^2,d=ar^3$
Next
a and b are the roots of $x^2 - 3x + p = 0$
So
$ab=p,a+b=3$
or $a.ar=p,a+ar=3$
or $a^2r=p,a(1+r)=3$ (1)
Also
c, d are roots of $x^2 - 12x + q = 0$
So
$cd=q,c+d=12$
or $ar^2.ar^3=p,ar^2+ar^3=12$
or $a^2r^5=p,ar^2(1+r)=12$ (2)
Dividing (2) by (1), we get
$r^2=4$
$r=2$
Using r=2, we get a=1
Now
$\frac{q+p}{q-p}=\frac{cd+ab}{cd-ab}$

or $\frac{q+p}{q-p}=\frac{a^2r^5+a^2r}{a^2r^5-a^2r}$

or $\frac{q+p}{q-p}=\frac{r^4+1}{r^4-1}$

or $\frac{q+p}{q-p}=\frac{17}{15}$
This completes the solution.
The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that $a:b=\left ( m+\sqrt{m^2-n^2} \right ) : \left ( m-\sqrt{m^2-n^2} \right )$
Solution 👉 Click Here

Solution
Given that two numbers are, so
$AM=\frac{a+b}{2}$
$GM=\sqrt{ab}$
Given that
$\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$
Now we have to show that
$\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}=\frac{a}{b}$

Using componendo dividendo rule, we get
$\frac{2m}{2\sqrt{m^2-n^2}}=\frac{a}{b}$

or $\frac{m}{\sqrt{m^2-n^2}}=\frac{a}{b}$

or $\frac{\left ( \frac{m}{n}\right )}{\sqrt{\left ( \frac{m}{n}\right )^2-1}}=\frac{a}{b}$

or $\frac{\left ( \frac{a+b}{2\sqrt{ab}}\right )}{\sqrt{\left ( \frac{a+b}{2\sqrt{ab}}\right )^2-1}}=\frac{a}{b}$

or $\frac{\left ( \frac{a+b}{2\sqrt{ab}}\right )}{\sqrt{\frac{(a+b)^2-4ab}{4ab} } }=\frac{a}{b}$

or $\frac{\left ( \frac{a+b}{2\sqrt{ab}}\right )}{\sqrt{\frac{(a-b)^2}{4ab} } }=\frac{a}{b}$

or $\frac{\left ( \frac{a+b}{2\sqrt{ab}}\right )}{\frac{a-b}{2\sqrt{ab}}}=\frac{a}{b}$

or $\frac{a+b}{a-b}=\frac{a}{b}$

Using componendo dividendo rule, we get
$\frac{a}{b}=\frac{a}{b}$

This completes the solution.
Find the sum of the following series up to n terms:
(i) 5 + 55 +555 + …
(ii) 0.6 +0.66 +0.666+…
Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.
Solution 👉 Click Here

Solution
Find the 20th term of the series is
(20th term of 2,4,6,..)*(20th term of 4,6,8,.. )
This completes the solution.
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
Bidhan buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Solution 👉 Click Here

Solution
The unpaid amount forms a sequence given as
18000, 17000,16000,...,1000 for 18 installments (18 terms)
1000, 2000,3000,...,18000 for 18 installments (18 terms)
Using formula, the sum of unpaid amounts is
$s_n=\frac{n}{2}(2a+(n-1)d)$
or $s_n=\frac{18}{2}(2*1000+17*1000)=1,71,000$
So, the interset paiid is
10% of 171000=17100
So, the cost of the scooter is
22000+17100=39100
This completes the solution.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Solution 👉 Click Here

Solution
Given that, one person writes letter to his 4 friends, aagain, each of these 4 friends will write letter to their 4 friends. So, the number of letters written here will be $4^2$ . This continues....
So, the number forms a GP given as
$4,4^2,4^3,...,4^8$
Here, a=4,r=4,n=8
So,
$S_n= \frac{a(r^n-1)}{r-1}$
or $S_8= \frac{4(4^8-1)}{4-1}$
or $s_8=87380$
Now, cost to mail the letters is
0.5*87380=Rs 43690
This completes the solution.
A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution 👉 Click Here

Solution
Let us assume that, 150 workers finish a work in x days.
Then, according to question
Number of workers working in each days are respectively
150,146,142,......and it continues up to (x+8) terms [sine 8 more days are required]
This shows that, the numbers are in A.P. with first term 150, common difference -4 and number of terms as (x+8)
So,
$150x= \frac{n}{2} [2a+(n-1)d]$
or $150x= \frac{x+8}{2} [2.150+(x+8-1)(-4)]$
orx=17 or x=-32
or x=17days can NOT be nagative
This completes the solution.

EXERCISE 8.2 [NCERT, page 145]

Find the 20th and nth terms of the G.P. $\frac{5}{2},\frac{5}{4},\frac{5}{8},...$
Solution 👉 Click Here

Given that the GP is
$\frac{5}{2},\frac{5}{4},\frac{5}{8},...$
The first term is
$t_1=a=\frac{5}{2}$
The common ratio is
$r=\frac{t_2}{t_1}=\frac{\frac{5}{4}}{\frac{5}{2}} =\frac{1}{2}$
Now, the 20th term is
$t_{20}=ar^{19}=\frac{5}{2}*(\frac{1}{2})^{19}=\frac{5}{1048576}$
Next, the nth term is
$t_n=ar^{n-1}=\frac{5}{2}*(\frac{1}{2})^{n-1}=\frac{5}{2^n}$
This completes the solution
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution 👉 Click Here

Given that the 8th term is
$t_{8}=ar^7=192$
The common ratio is
$r=2$
So, the using 8th term, we get
$t_8=192$
or $ar^7=192$
or $a.2^7=192$
or $a=\frac{3}{2}$
Now, the 12th term is
$t_{12}=ar^{11}=\frac{3}{2}*(2^{11})=3072$
This completes the solution
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that $q^2 = ps$ .
Solution 👉 Click Here

Given that
$t_{5}=ar^4=p$
$t_{8}=ar^7=q$
$t_{11}=ar^{10}=s$
By dividing, we get
$r^3=\frac{t_8}{t_5}=\frac{q}{p}$
$r^3=\frac{t_{11}}{t_8}=\frac{s}{q}$
So, the using both $r^3$ , we get
$r^3=r^3$
or $\frac{q}{p}=\frac{s}{q}$
or $q^2 = ps$
This completes the solution
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.
Solution 👉 Click Here

Given that
$t_4=(t_2)^2$
or $ar^3=(ar)^2$
or $ar^3=a^2r^2$
or $a=r$
Also, given that
$a=-3$
Now, 7th term is
$t_7=ar^6$
or $t_7=r^7$
or $t_7=(-3)^7$
or $t_7=-2187$
This completes the solution
Which term of the following sequences:
1. $2,2\sqrt{2},4,...$ is 128 ?
  Solution 👉 Click Here
  
  Given that
  $2,2\sqrt{2},4,...$
  The first term is
  $t_1=a=2$
  The common ratio is
  $r=\frac{t_2}{t_1}=\frac{2}{2\sqrt{2}} =\sqrt{2}$
  The given term is
  $t_n=128$
  or $ar^{n-1}=128$
  or $2*(\sqrt{2})^{n-1}=128$
  or $(\sqrt{2})^{n-1}=64$
  or $(\sqrt{2})^{n-1}=(\sqrt{2})^{12}$
  or $n-1=12$
  or $n=13$
  This completes the solution
2. $\sqrt{3},3,3\sqrt{3},...$ is 729 ?
  Solution 👉 Click Here
  
  Given that
  $\sqrt{3},3,3\sqrt{3},...$
  The first term is
  $t_1=a=\sqrt{3}$
  The common ratio is
  $r=\frac{t_2}{t_1}=\frac{3}{\sqrt{3}} =\sqrt{3}$
  The given term is
  $t_n=729$
  or $ar^{n-1}=729$
  or $\sqrt{3}*(\sqrt{3})^{n-1}=729$
  or $(\sqrt{3})^n=(\sqrt{3})^{12}$
  or $n=12$
  This completes the solution
3. $\frac{1}{3},\frac{1}{9},\frac{1}{27},...$ is $\frac{1}{19683}$
  Solution 👉 Click Here
  
  Given that
  $\frac{1}{3},\frac{1}{9},\frac{1}{27},...$ is $\frac{1}{19683}$
  The first term is
  $t_1=a=\frac{1}{3}$
  The common ratio is
  $r=\frac{t_2}{t_1}=\frac{\frac{1}{3}}{\frac{1}{9}} =\frac{1}{3}$
  The given term is
  $t_n=\frac{1}{19683}$
  or $ar^{n-1}=\frac{1}{19683}$
  or $\frac{1}{3}*(\frac{1}{3})^{n-1}=\frac{1}{19683}$
  or $(\frac{1}{3})^n=(\frac{1}{3})^9$
  or $n=9$
  This completes the solution
For what values of x, the numbers $\frac{2}{7},x,\frac{7}{2}$ are in G.P.?
Solution 👉 Click Here

Given that
$\frac{2}{7},x,\frac{7}{2}$
We know that, three consecutive terms a,b,c are in GP if $b^2=ac$
So, for three numbers $\frac{2}{7},x,\frac{7}{2}$ , we can write that
$x^2=\frac{2}{7}*\frac{7}{2}$
or $x=1$
This completes the solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

0.15, 0.015, 0.0015, ... 20 terms
Solution 👉 Click Here

Given that
0.15, 0.015, 0.0015, ...
The first term is
$t_1=a=0.15$
The common ratio is
$r=\frac{t_2}{t_1}=\frac{0.015}{0.15} =0.1$
We have to cimpute the sum up to 20 terms
So,
n=20
Now, sum of 20terms is
$S_n=\frac{a(r^n-1)}{r-1}$
or $S_n=\frac{0.15((0.1)^{20}-1)}{0.1-1}$
or $S_n=\frac{0.15(1-(0.1)^{20})}{0.9}$
or $x=\frac{1}{6} (1-(0.1)^{20})$
This completes the solution
$\sqrt{7},\sqrt{21},3\sqrt{7},...$ n terms
Solution 👉 Click Here

Given that
$\sqrt{7},\sqrt{21},3\sqrt{7},...$
The first term is
$t_1=a=\sqrt{7}$
The common ratio is
$r=\frac{t_2}{t_1}=\frac{\sqrt{21}}{\sqrt{7}} =\sqrt{3}$
We have to cimpute the sum up to n terms
Now, sum of nterms is
$S_n=\frac{a(r^n-1)}{r-1}$
or $S_n=\frac{\sqrt{7}((\sqrt{3})^n-1)}{\sqrt{3}-1}$
or $S_n=[\sqrt{7}(\sqrt{3}+1)]\frac{\sqrt{3}^n-1}{2}$
or $S_n=[\sqrt{7}(\sqrt{3}+1)]\frac{3^{\frac{n}{2}}-1}{2}$
This completes the solution
$1,-a,a^2,-a^3,...$ n terms (if a≠-1)
Solution 👉 Click Here

Given that
$1,-a,a^2,-a^3,...$
The first term is
$t_1=a=1$
The common ratio is
$r=\frac{t_2}{t_1}=\frac{-a}{1} =-a$
We have to cimpute the sum up to n terms
Now, sum of nterms is
$S_n=\frac{a(r^n-1)}{r-1}$
or $S_n=\frac{1((-a)^n-1)}{-a-1}$
or $S_n=\frac{1-(-a)^n}{a+1}$
This completes the solution
$x^3,x^5,x^7,...$ n terms (if $n \ne \pm$ 1)
Solution 👉 Click Here

Given that
$x^3,x^5,x^7,...$
The first term is
$t_1=a=x^3$
The common ratio is
$r=\frac{t_2}{t_1}=\frac{x^5}{x^3} =x^2$
We have to cimpute the sum up to n terms
Now, sum of nterms is
$S_n=\frac{a(r^n-1)}{r-1}$
or $S_n=\frac{x^3((x^2)^n-1)}{x^2-1}$
or $S_n=\frac{x^3(x^{2n}-1)}{x^2-1}$
This completes the solution
Evaluate $\displaystyle \sum_{k=1}^{11} (2+3^k)$
Solution 👉 Click Here

Given that
$\displaystyle \sum_{k=1}^{11} (2+3^k)$
= $\displaystyle \sum_{k=1}^{11} 2+\sum_{k=1}^{11} 3^k$
= $\displaystyle 11.2+\sum_{k=1}^{11} 3^k$
= $22+ (3^1+3^2+3^3+...+3^{11})$
= $22+ \frac{a(r^n-1)}{r-1}$ where a=3, r=3,n=11
= $22+ \frac{3(3^{11}-1)}{3-1}$
= $22+ \frac{3}{2} (177146)$
This completes the solution
The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.
How many terms of G.P. $3, 3^2, 3^3$ , … are needed to give the sum 120?
The sum of first three terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Given a G.P. with a = 729 and 7th term 64, determine $S_7$ .
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and $128, 32, 8, 2,\frac{1}{2}$
Show that the products of the corresponding terms of the sequences $a, ar, ar^2,…ar^{n - 1}$ and $A, AR, AR^2, … AR^{n - 1}$ form a G.P, and find the common ratio.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that $a^{q - r} b^{r - p}c^{p - q} = 1$ .
Solution 👉 Click Here

Let A be the first term and R be the common ration , then
$t_p=AR^{p-1}=a$
$t_q=AR^{q-1}=b$
$t_r=AR^{q-1}=c$
Now,
$a^{q - r} b^{r - p}c^{p - q}$
or $(AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}. (AR^{q-1})^{p-q}$
or $A^0.R^0$
or1
This completes the solution
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that $p^2 = (ab)^n$ .
Solution 👉 Click Here

Let A be the first term and R be the common ration , then
$t_1=A=a$
$t_n=AR^{n-1}=b$
Now,
$ab=A.AR^{n-1}$
or $ab= A^2R^{n-1}$
Next
$P=A*AR*AR^2*...*AR^{n-1}$
or $P=A^n* (R*R^2*...R^{n-1})$
or $P=A^n* R^{\frac{n(n-1)}{2}}$
or $P^2=A^{2n}* R^{n(n-1)}$
or $P^2=(A^2* R^{n-1})^n$
or $P^2=(ab)^n$
This completes the solution
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $\frac{1}{r^n}$
If a, b, c and d are in G.P. show that $(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2$ .
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Find the value of n so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between a and b.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $\left ( 3+2\sqrt{2} \right ): \left ( 3-2\sqrt{2}\right )$
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A \pm \sqrt{(A+G)(A-G)}$
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Linear Programming

Posted on January 7, 2024January 7, 2024 by MEAN

LPP is a technique of mathematical modeling to solve real life problems whose requirements are given in linear relations. It helps to find “best” solution (minimum and maximum) under given conditions.
It is a type of optimization problem that determines the feasible region (a region that contains all the possible solutions of an LPP) and optimizes the solution to get the maximum or minimum function value.
In simple terms, a linear programming problem is a set of linear equations that are used to find the value of variables to optimize the objective functions. LPP can be solved in two different ways.

LPP (linear programming problem) वास्तविक जीवनका समस्याहरू समाधान गर्ने एक गणितीय मोडलिङ हो जसको शर्तहरु रेखिय सम्बन्धमा दिएको हुन्छ । यसले दिइएको अवस्थाहरूमा “उत्तम” समाधान (न्यूनतम वा अधिकतम) को बारेमा थाहा पाउन मद्दत गर्दछ।
यो एक प्रकारको अप्टिमाइजेसन समस्या हो जसले सम्भाव्य क्षेत्र (सबै सम्भावित समाधानहरू समावेश गर्ने क्षेत्र) निर्धारण गर्छ र अधिकतम वा न्यूनतम मान प्राप्त गर्न मद्दत गर्छ ।
समान्य भाडषामा, LPP (linear programming problem) भनेको रेखिय समीकरणहरूको समुह हो जसबाट “उत्तम” समाधान (न्यूनतम वा अधिकतम) को खोज गरिन्छ।

Linear Programming Applications_Example 1

Let us take a real-life problem to understand linear programming.
A company produces two types of pots X and Y by using copper and steel. Type X pot requires 300 g of copper and 100 g of steel while Y requires 100 gram of copper and 200 grams of steel. The X pot brings profit of Rs. 400 and the Y pot brings profit of Rs. 500. Find the number of units of each type of pots that the company should produce with 5kg copper and 12kg of steel to achieve maximum profit.
Given the situation, let us take up different scenarios to analyse how the profit can be maximized.

The company can decide to produces X type pots, in this case, he can create 5000/100= 50 copper slot, 12000/200=60 steel slot. This would give him to produce 50 pots with a profit of Rs 500×50 = Rs 12500.
The company can decide to produces Y type pots, in this case, he can create 5000/300= 16 copper slot, 12000/100=120 steel slot. This would give him to produce 16 pots with a profit of Rs 400×14 = Rs 6400.
Similarly, there can be many strategies which the company can devise to maximize his profit by allocating the available row materials to the two types of pots. We do a mathematical formulation of the discussed LPP to find out the strategy which would lead to maximum profit.

Linear Programming Applications_Example 2

A home decorator company received an order to manufacture tables. The first consignment requires up to 50 tables. There are two types of tables, The first type requires 15 hours of the labour force (per piece) to be constructed and gives a profit of Rs 5000 per piece to the company. Whereas, the second type requires 9 hours of the labour force and makes a profit of Rs 3000 per piece. However, the company has only 540 hours of workforce available for the manufacture of the tables. With this information given, you are required to find a deal which gives the maximum profit to the company.

Given the situation, let us take up different scenarios to analyse how the profit can be maximized.

He decides to construct all the tablesof the first type. In this case, he can create 540/15 = 36 tables. This would give him a profit of Rs 5000 × 36 = Rs 180,000.
He decides to construct all the tables of the second type. In this case, he can create 540/9 = 60 tables. But the first consignment requires only up to 50 cabinets. Hence, he can make profit of Rs 3000 × 50 = Rs 150,000.
He decides to make 15 tables of type 1 and 35 of type 2. In this case, his profit is (5000 × 15 + 3000 × 35) Rs 180,000.

Similarly, there can be many strategies which he can devise to maximize his profit by allocating the different amount of labour force to the two types of tables. We do a mathematical formulation of the discussed LPP to find out the strategy which would lead to maximum profit.

Basic terminologies of LPP

Decision Variable: These are quantities to be determined.
Objective Function: The function that need to optimized.
Constraints: Linear equations that represents decision variables on how to use the available resource (for example, amount of time, number of people, etc.)
Feasible Solution: Set all the possible solutions that satisfy the given constants.
Optimal Solution: It is the best possible solution among all the possible feasible solutions.
Slack Variables : Slack variable represents an unused quaintly of resources ; it is added to less than or equal (<) to type constraints in order to get an equality constraint.
Surplus Variables : A surplus variable represents the amount by which solution values exceed a resource. These variables are also called ‘Negative Slack Variables’. Surplus variables carry a zero coefficient in the objective function. it is added to greater than or equal to (>) type constraints in order to get an equality constraint.

Characteristics of LPP

Decision variables will decide the output.
The objective function should be specified quantitatively.
Constraints (limitations) should be expressed in mathematical form.
Relationships between two or more variables should be linear.
The values of the variables should always be non-negative or zero.

Methods of Solving LPP

LPP बिभिन्न तरिकाबाट हल गर्न सकिन्छ।

Graphical method
Simplex method
North West Corner Method
Least Square Methods

Graphical Methods of Solving LPP

LPP models can be solved by graphical method if two variable are presented in the model.
The general process while solving LPP by graphical methods, are given below

graph the inequalities (constraints)
form walled-off region (feasible region)
find the corner points in feasible region
test corner points in the object function
find the solution (maximum or minimum)

Simplex Methods of Solving LPP

LPP models can be solved by simplex method if two or more variable are presented in the model.
The general process while solving LPP by simplex methods, are given below

Convert subjective function with ≤ inequality (standard form)
Introduce slack variables to convert inequality into equation
Prepare simplex table
Find pivot element [column (negative, lowest), Find pivot row (positive, lowest)]
Convert pivot element to 1 and else to zero
If optimal solution is NOT reached, iterate the process from 4 onwards

A company produces two types of pots X and Y by using copper and steel. Type X pot requires 300 g of copper and 100 g of steel while Y requires 100 gram of copper and 200 grams of steel. The X pot brings profit of Rs. 400 and the Y pot brings profit of Rs. 500. Find the number of units of each type of pots that the company should produce with 5kg copper and 12kg of steel to achieve maximum profit.

Solution

The decision variables: Since the question has asked for an optimum number of pots, that’s what our decision variables is. Let us say
x = number of X pots produced
y = number of Y pots produced
The constraints: Since the company can’t produce a negative number of pots, a natural constraint is:
x ≥ 0
y ≥ 0

The subjective function

pots	copper	steel	Profit
X	300	100	400
Y	100	200	500
Availability	5000	12000	?

Given
Each unit of X requires 300 unit of copper and 100 units of steel
Each unit of Y requires 100 unit of copper and 200 units of steel
The company has a total of 5 kg (5000 g) of copper and 12 kg (12000 g) of steel.
Thus, the subjective functions are
300x+100y≤5000
100x+200y≤12000

The objective function
We need to optimize the Profit. On each sale, the company makes a profit of
Rs 400 per unit X sold.
Rs 500 per unit Y sold.
The Profit Function is:
Z =400x + 500y
LPP modeling
Therefore, the LPP model is (in 100) is
Maximize Z = 4x + 5y, subject to:
3x+1y≤50
1x+2y≤120
x ≥ 0
y ≥ 0

Solution

Step 1: graph the inequalities (constraints)
We plot the following inequalities in a graph.

$2x + y \le 100$
Let us take a table for plot points

x	0	50
y	100	0

The graph is given below.

$x + y \le 80$
Let us take a table for plot points

x	0	80
y	80	0

The graph is shown with a label.

$x \ge 0,x \ge 0$
The graph is shown with a label.

Step 2: form a walled-off region (feasible region)
In this LPP, the feasible region is a polygon OABC labeled in the graph.

Step 3: find the corner points in the feasible region
In this LPP, the corner points of polygon OABC are O = (0,0), A = (16.5,0), D= (0,50)
NOTE: The corner point are obtained by solving two equations, which do intersect

Step 4: test corner points in the object function

Objective function	Point	Value	Remarks
Z=4x+5y	(0,0)	$Z=4\times 0+5\times 0=0$	Minimum
Z=4x+5y	(16.5,0)	$Z=4\times 16.5+5\times 0=66$
Z=4x+5y	(0,50)	$Z=4\times 0+5\times 50=250$	Maximum

Step 5: find the solution (maximum or minimum)
The maximum value of the function Z = 4x + 5y is 250 at the point (0,50)
The minimum value of the function Z = 4x + 5y is 0 at the point (0,0).
The graphical solution of this modeling can be shown below.

In the LPP, the graphical method is very useful in lower dimensions. If we have more than two variables then we’re looking at some higher dimensional shape, then we can’t interpret anything easily in three dimensions visually. Therefore, the simplex method gives an algorithm for solving these problems with any number of variables. Graphical method can be applied in 2D. In 3D and higher dimensions+, identifying an optimal solution using the graphical method is no longer feasible.So, simplex method can be applied to 1D, 2D, 3D and 3D+ linear programs. In other words, simplex method can be used for theoretically unlimited amounts of optimization variables

The general process for solving the maximization LPP model by the simplex method is given below

Convert subjective function with $\le$ inequality (standard form)
Introduce slack variables to convert inequality into an equation
Prepare simplex table
Find the pivot column (negative, lowest)
Find pivot row (positive, lowest)
Find pivot element
Convert pivot element to 1 and else to zero
Iterate the process from 4 onwards

Test your understandings

Why do we make an objective function with a negative coefficient?
The initial simplex table represents an augmented matrix, where slack variables form an identity matrix.

In an LPP model
Maximize $z = 400x + 500y$ , subject to:
$300x+100y \le 5000$
$100X+200y \le 12000$

x	y	u	v	z
300	100	1	0	0	5000
100	200	0	1	0	12000
-400	-500	0	0	1	0

the objective function is z=400x+500y, which is in the linear form written as z-400x-500y=0 in order to produce a unit matrix as shown in the table above

Why do we choose the most negative entry in the bottom row?

We know that the simplex method begins at a corner point where all the main variables are zero. It then moves from a corner point to the adjacent corner point always increasing the value of the objective function.
when we choose the most negative entry in the bottom row, we are trying to increase the value of the objective function ,
for example, the coefficient of y (which is 500) , this entry will increase the value of the objective function the quickest

Why do we find quotients, and why does the smallest quotient identify a row?
When we choose the most negative entry in the bottom row, we are trying to increase the value of the objective function by bringing in the variable, say y [in the example below]. But we cannot choose any value for y. Therefore, using the lowest quotients guarantees that we do not violate the constraints and that the effect is high to increase the value of the objective function.

x	y	u	v	z		ratio
300	100	1	0	0	5000	5000/100=50
100	200	0	1	0	12000	12000/200=60
-400	-500	0	0	1	0

For example,

in a model, Maximize $z = 400x + 500y$ , subject to:
$300𝑥+100𝑦 \le 5000$
$100x+200𝑦 \le 12000$
the smallest ratio is 50, which ensures the validity of the constraints

Why do we identify the pivot element?
The simplex method begins with a corner point and then moves to the next corner point to improve the value of the objective function. Therefore, the value of the objective function is improved by changing the number of units of the variables. We may add the number of units of one variable while throwing away the units of another. Pivoting allows us to do just that and help us to utilize available resources

x	y	u	v	z		ratio
300	100	1	0	0	5000	5000/100=50
100	200	0	1	0	12000	12000/200=60
-400	-500	0	0	1	0

For example,

in a model, Maximize $z = 400x + 500y$ , subject to:
$300𝑥+100𝑦 \le 5000$
$100x+200𝑦 \le 12000$

the value of the objective function is improved by changing the number 100[a12] to 1 and else to zero

Why are we finished when there are no negative entries in the bottom row?
Since all variables are non-negative, the highest value that can be achieved is guaranteed.

Example 1

Find the maximum value of the following LPP problem.
Maximize: Z = 50x + 18y
Subject to: $2x + y \le 100, x + y \le 80, x \ge 0 , y\ge 0$

Example 2

Find the maximum value of the following LPP problem.
Maximize: Z = 25x + 40y
Subject to: $2x + y \le 10, x + 2y \le 6, x \ge 0 , y\ge 0$

Solution

Step 1: Since all the subjective function are with $\le$ inequality, Prepare a simplex Tableaux

x	y	u	v
2	1	1	0	10
1	2	0	1	6
-25	-40	0	0	0

Step 2: Find the pivot column, pivot row, and pivot element.
(a) Largest negative entry is -40, thus $C_2$ is the pivot column.
(b) Smallest positive ratio is 3, thus $R_2$ is the pivot row.
(c) Based on (a) and (b), 2 is the pivot element.

x	y	u	v			ratio
2	1	1	0	10		$\frac{10}{1}=10$
1	2	0	1	6	⇐	$\frac{6}{2}=3$
-25	-40	0	0	0
	⇑

Step 3: Operating Row equivalent operation, convert pivot element 2 to 1, and else to 0.

x	y	u	v		Operation	Stage
3/2	0	1	-1/2	7	$R_1 \approx R_1-R_2$	2
1/2	1	0	1/2	3	$R_2 \approx \frac{1}{2} R_2$	1
-5	0	0	20	120	$R_3 \approx R_3+40 R_2$	3

Step 4: Since a negative entry is found in row 3, again iterate the process to find the pivot column, pivot row, and pivot element.
(a) Largest negative entry is -5, thus $C_1$ is the pivot column.
(b) Smallest positive ratio is 14/3, thus $R_1$ is the pivot row.
(c) Based on (a) and (b), $3/2$ is the pivot element.

x	y	u	v			ratio
3/2	0	1	-1/2	7	⇐	$\frac{7}{3/2}=14/3$
1/2	1	0	1/2	3		$\frac{3}{1/2}=6$
-5	0	0	20	120
⇑

Step 5: Operating Row equivalent operation, convert pivot element 3/2 to 1, and else to 0.

x	y	u	v		Operation	Stage
1	0	2/3	-1/3	14/3	$R_1 \approx \frac{2}{3}R_1$	1
0	1	-1/3	2/3	2/3	$R_2 \approx R_2-\frac{1}{2} R_1$	2
0	0	10/3	55/3	430	$R_3 \approx R_3+5 R_1$	3

Step 6: Since all entries in row 3 are non-negative, the solution is achieved. Here, the maximum value is
$Z=\frac{430}{3}$ at x=14/3,y=2/3

Question

Find the maximum value of the following LPP problem.
Maximize: P=x+2y+z
Subject to: $x-y+2z\le 10, 2x+y+3z \le 12, x \ge 0 , y\ge 0,z\ge 0$

Complex Number

Posted on January 7, 2024January 7, 2024 by MEAN

Introduction

A complex number is an extended version of real number in the form
x + iy; 𝑥∈ℝ
Euler (1707 – 1783) introduced the imaginary unit ‘i’ (read as iota) for √-1 with property
$i^2+ 1 = 0$
Therefore, imaginary unit i is the solution of an equation
$x^2+ 1 = 0$

Acomplex is written in the STANDARD form as z = x+iy where x and y are real numbers

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Note

The cartesian form of complex number can be written z=x+iy [standard form] or z=(x,y) [order pair form]
In z = (x,y,), x is real part denoted as Re (z), and y is imaginary part denoted as Im (z)
In z = (x,y), Re (z) = x and Im (z) =y
A complex number z = x+iy, is purely real if y = 0 i.e. Re (z) = 0, and purely imaginary if x = 0 i.e. Im (z) = 0.
A complex number z = x+iy, is zero if x = y = 0 i.e. Re (z) = Im (z) = 0

The introduction of complex numbers gives rise to the fundamental theorem of algebra. In the 16th century, Italian mathematician Gerolamo Cardano used complex numbers to find solutions to cubic equations.

Then Italian mathematician Rafael Bombelli developed the rules for addition, subtraction, multiplication, and division of complex numbers. A more abstract formalism for complex numbers was developed by the Irish mathematician William Rowan Hamilton.

Meaning of i

In the complex number system, i is called imaginary unit. Tthe value of i is (0, 1). Thus, we can write i=(0,1).

If we expand a complex number z=(x,y) as z=(1,0)x+(0,1)y then (1,0) is unit of real part denoted by 1 and (0,1) is unit of imaginary part and denoted by i. Here i is imaginary unit with
$i^2=-1$ .
A complex number is visually represented in Argand diagram. Here i is operator giving anticlockwise quarter turn such that $i^2=-1$ .

NOTE

We should NOT mean i as non-existence number nor a number that exist only in imagination
i is a number that denotes imaginary unit (0,1)
i is an anticlockwise quarter turn operator for (x,y), thus
$i^2=(0,1) (0,1) = i (0,1) =(-1,0) =-1 (1,0)=-1$

Positive powers of i

In general, for any integer k,
$i^{4k} = 1, i^{4k+1} = i, i^{4k+2} = -1, i^{4k+3} = -i$ .
For example:
$i^{23} = i^{4.5+3} = (i^4)5. i ^3= 1^5. (-i) = -i.$

The verification on the positive powers of i are as follows.
$i^2 = -1$
$i^3 = i2. I = (-1).i = - i$
$i^4 = (i^2)^2 = (-1)^2 = 1$
$i^5 = i^{4 + 1}= i^4. i = 1. i = i$
$i^6 = i^{4 + 2} = i^4. i^2= 1. 1 = 1$ and so on

Absolute Value

In a complex number
$z= x + iy$
The absolute value is Modulus, a non-negative real number denoted by
| z | and defined by
$|𝑧|=\sqrt{x^2+y^2}$
Geometrically,
|𝑧| is distance of z from origin

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Note
Due to order property, z₁ <z₂ is meaningless unless z₁ and z₂ both are real.
However,| z₁ | <| z₂ | means z₁is closer than z₂.
Distance between $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is denoted by
$| z_1 -z_2 |$
and defined by
$| z_1 -z_2 |=\sqrt{( x_1 -x_2 )^2 +( y_1 -y_2 )^2 }$

NOTE:
$( x+iy )^2 =( x+iy )( x-iy )$

Conjugate

The conjugate of a complex number z=x+iy is denoted $\bar{z}$ and defined by
$\bar{z}=x-iy$
Geometrically, $\bar{z}$ is reflection of $z$ on real axis

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Properties of Conjugate numbers

If $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ then $\overline{z_1 +z_2 }=\bar{z_1} +\bar{z_2}$
Proof
$\overline{z_1 +z_2 }=\overline{( x_1 +iy_1 )+( x_2 +iy_2 )}$
or $\overline{z_1 +z_2 }=\overline{( x_1 +x_2 )+i( y_1 +y_2 )}$
or $\overline{z_1 +z_2 }=( x_1 +x_2 )-i( y_1 +y_2 )$
or $\overline{z_1 +z_2 }=( x_1 -iy_1 )+( x_2 -iy_2 )$
or $\overline{z_1 +z_2 }=\bar{z}_1 +\bar{z}_2$
The sum of a complex number z and its conjugate $\bar{z}$ is twice of its real part
Proof
If $z=x+iy$ be a complex number then $\bar{z}=x-iy$ , where
$z+\bar{z}=2x$
or $x=\frac{z+\bar{z}}{2}$
or $Rez=\frac{z+\bar{z}}{2}$
The difference of z and its conjugate $\bar{z}$ is twice of its imaginary part
Proof
If z=x+iy be a complex number then $\bar{z}=x-iy$ , where
$z-\bar{z}=2iy$
or $y=\frac{z-\bar{z}}{2i}$
or $Imz=\frac{z-\bar{z}}{2i}$
The real and imaginary parts of a complex number can be extracted using its conjugate

Theorem: An important property

Let z=x+iy be a complex number then $z.\bar{z}=|z|^2$

Proof
Given z=x+iy be a complex number, then $\bar{z}=xiy$ .
Now,
$z.\bar{z}=( x+iy )( x-iy )$
or $z.\bar{z}=( x )^2 -( iy )^2$
or $z.\bar{z}=x^2 +y^2$ (i)
Also
$| z |=\sqrt{x^2 +y^2}$
or $|z|^2 =x^2 +y^2$ (ii)
From (i) and (ii), we get
$z.\bar{z}=|z|^2$

Question

Theorem: Triangle inequality: $| z_1 +z_2 |\le | z_1 |+| z_2 |$

Proof>br> We know that
$| z_1 +z_2 |^2 =( z_1 +z_2 )( \overline{z_1 +z_2 } )$
or $| z_1 +z_2 |^2 =( z_1 +z_2 )( \bar{z}_1 +\bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +z_2 \bar{z}_1 +z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\bar{\bar{z}}_2 \bar{z}_1 +z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\overline{z_1 \bar{z}_2 }+z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =z_1 \bar{z}_1 +2Re( z_1 \bar{z}_2 )+z_2 \bar{z}_2$
or $| z_1 +z_2 |^2 \le z_1 \bar{z}_1 +2| z_1 \bar{z}_2 |+z_2 \bar{z}_2$
or $| z_1 +z_2 |^2 =|z_1|^2 +2| z_1 || \bar{z}_2 |+|z_2|^2$
or $| z_1 +z_2 |^2 =( | z_1 |+| z_2 | )^2$
or $| z_1 +z_2 |\le | z_1 |+| z_2 |$
This completes the proof.

Corollary

$| z_1 +z_2 |\ge | z_1 |-| z_2 |$

Proof
or $| z_1 |=| z_1 +z_2 +( -z_2 ) |$
or $\le | z_1 +z_2 |+| -z_2 |$
or $=| z_1 +z_2 |+| z_2 |$
Thus, $| z_1 +z_2 |\ge | z_1 |-| z_2 |$
$| z_1 -z_2 |\le | z_1 |+| z_2 |$

Proof
$| z_1 +z_2 | \le | z_1 |+| z_2 |$
Now, replacing $z_2$ by $-z_2$ we get
$| z_1 -z_2 |\le | z_1 |+| -z_2 |$
Thus, $| z_1 -z_2 |\le | z_1 |+| z_2 |$
$| z_1 -z_2 |\ge | z_1 |-| z_2 |$

Proof
$| z_1 +z_2 | \ge | z_1 |-| z_2 |$
Now, replacing $z_2$ by $-z_2$ we get
$| z_1 -z_2 |\ge | z_1 |-| -z_2 |$
Thus, $| z_1 -z_2 |\ge | z_1 |-| z_2 |$

Algebra of complex number

Complex plane looks like an ordinary two-dimensional plane of z=( x,y ), but z=( x,y ) is a single number losing order axioms. Fundamental operations on complex number are defined as below.

Equality
Two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ are equal if
$x_1 =x_2 \text{ and } y_1 =y_2$
Addition
Sum of two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is defined as
$z_1 +z_2 =( x_1 +iy_1 )+( x_2 +iy_2 )=( x_1 +x_2 )+i( y_1 +y_2 )$
According to definition, $z_1 +z_2$ corresponds to resultant vector addition.
Multiplication
Product/multiplication of two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is defined as
$z_1 .z_2 =( x_1 +iy_1 ).( x_1 +iy_2 )$
or $z_1 .z_2 =x_1 x_2 +ix_1 y_2 +ix_2 y_1 +{{i}^{2}}y_1 y_2$
or $z_1 .z_2 =( x_1 x_2 -y_1 y_2 )+i( x_1 y_2 +x_2 y_1 )$
The product is neither scalar nor the vector product of ordinary vector analysis. This departure is due to $i^2=-1$

Also, complex number C is a field. Thus, the complex number satisfy all Field axioms as below.

Closer: $\forall a,b \in G \implies a \ast b \in G$
Additive associative
$\forall a,b,c \in G \implies (a+ b) + c=a + (b + c)$
Additive identity
$\forall a \in G \implies a \ast e =e \ast a =a$
Additive identity (0,0)
Additive Inverse
$\forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e$
Additive inverse of z=x+iy is -z=-x-iy
Additive commutative
$\forall a,b \in G \implies a \ast b =b \ast a$
Distributive
$\forall a,b,c \in G \implies (a + b)\bullet c =a c+bc$
Multiplicative associative
$\forall a,b,c \in G \implies (a. b) . c=a . (b . c)$
Multiplicative identity
$\forall a \in G \implies a \ast e =e \ast a =a$
Multiplicative identity (1,0)
Multiplicative inverse
$\forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e$
Multiplicative inverse of non-zero complex number z=x+iy is
$z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}$
Proof
Let z=x+iy be non-zero complex number and $z^{-1} =u+iv$ be its multiplicative inverse, then
$zz^{-1} =( 1,0 )$
or $( x+iy )( u+iv )=( 1,0 )$
or $( xu-yv )+i( xv+uy )=( 1,0 )$
Comparing real and imaginary parts separately, we get
ux-vy=1 and uy+vx=0
Solving for u and v , we get
$u=\frac{x}{x^2 +y^2}$ and $v=\frac{-y}{x^2 +y^2}$

Hence,
$z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}$
Multiplicative commutative
$\forall a,b \in G \implies a \ast b =b \ast a$

Algebric Structure

Axioms	Structure	Example
1-2	Semi Group	Example
1-3	Monoid	Example
1-4	Group	Example
1-5	Abelian Group	Example
1-6	Ring	Example
1-7	Assocoative Ring	Example
1-8	Assocoative Ring with Unity	Example
1-9	Division Ring (Skew Field)	Example
1-10	Field	Example

Polar form

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Let $z=x+iy$ be a complex number with magnitude r and amplitude $\theta$ , then
$x=r\cos \theta$ and $y=r\sin \theta$
Hence,
$z=x+iy=r\cos \theta +ir \sin \theta$
or $z=r( \cos \theta +i\sin \theta )$
Thus, a complex number z=x+iy is represented by polar coordinate $(r,\theta)$ , as
$z=r \cos \theta +i\sin \theta$
Here,
r is the length of z , and $\theta$ is argument of z with
$r=| z |=\sqrt{x^2 +y^2}$ , and $\theta =argz=\tan^{-1}( \frac{y}{x} )$

Example

Find polar form of complex number z=-5+5i

Solution
Given that
z=-5+5i
or z=5( -1+i)
or $z=5\sqrt{2}( -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i )$
or $z=5\sqrt{2}( \cos \frac{3\pi }{4}+\sin \frac{3\pi }{4}i )$

The relevance of complex number in polar form is that multiplication and division are simpler with this form than the Cartesian form.

Let $z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )$ and $z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )$ be two complex numbers, then
$z_1 z_2 =r_1 r_2 (\cos (\theta_1 +\theta_2 )+i\sin (\theta_1 +\theta_2 ))$
Proof
Given that $z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )$ and $z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )$ . Thus
$z_1 z_2 = r_1 (\cos \theta_1 +i\sin \theta_1 ) \times r_2 (\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 (\cos \theta_1 +i\sin \theta_1 )(\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 [\cos \theta_1 (\cos \theta_2 +i\sin \theta_2 )+i\sin \theta_1 (\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 [\cos \theta_1 \cos \theta_2 +i\cos \theta_1\sin \theta_2 +i\sin \theta_1\cos \theta_2 -\cos \theta_1\sin \theta_2 ]$
or $z_1 z_2 = r_1 r_2 [(\cos \theta_1 \cos \theta_2-\cos \theta_1\sin \theta_2 ) +i(\cos \theta_1\sin \theta_2 +\sin \theta_1\cos \theta_2) ]$
or $z_1 z_2 = r_1 r_2 [\cos (\theta_1 +\theta_2) +i \sin (\theta_1+ \theta_2)]$

Theorem

$arg( z_1 z_2 )=arg z_1 +argz_2$

Proof

Let $z_1 =r_1 ( \cos \theta_1 +i \sin \theta_1 )$ and $z_2 =r_2 ( \cos \theta_2 +i\sin \theta_2 )$ then
$z_1 .z_2=r_1 .r_2 [ \cos ( \theta_1 +\theta_2 )+i\sin ( \theta_1 +\theta_2 ) ]$
Thus,
$arg( z_1 z_2 )=argz_1 +argz_2$

Note:
Any complex number z has infinite arguments; all differ by multiple of $2\pi$ . The principal value is in the interval $[ -\pi ,\pi ]$

Some important property

Let $z_1 \text{ and } z_2$ be two complex number then $arg( z_1 .z_2 )=argz_1 +argz_2$
The argument of product of two complex number is sum of their arguments.
Proof
Let $z_1 =r_1 e^{ i\theta_1}$ and $z_2 =r_2 e^{ i\theta_2}$ then
$z_1 z_2 =( r_1 e^{ i\theta_1} )( r_2 e^{ i\theta_2} )$
or $z_1 z_2 =( r_1 r_2 )e^{i( \theta_1 +\theta_2 )}$
Hence,
$arg( z_1 .z_2 )=\theta_1 +\theta_2 =argz_1 +argz_2$
Let $z_1 \text{ and }z_2$ be two complex number then,
$arg( \frac{z_1 }{z_2 } )=argz_1 -argz_2$

The argument of quotient of two complex number is difference of their arguments.
Argument of complex number of the form $z=a+i.0, a > 0$ is 0
Argument of complex number of the form $z=a+i.0, a < 0$ is $\pi$
Argument of complex number of the form $z=0+i.b, b > 0$ is $\frac{\pi }{2}$
Argument of complex number of the form $z=0+i.b, b < 0$ is $-\frac{\pi }{2}$

Drag the point for intaraction : 👇

If $z =r( \cos \theta +i\sin \theta )$ then show that $z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )$

Question

Let $z=r( \cos \theta +i\sin \theta )$ be a complex number and its inverse is $z^{-1} =R( \cos \phi +i\sin \phi )$ then
$zz^{-1} =1$
or $r( \cos \theta +i\sin \theta )R( \cos \phi +i\sin \phi )=( 1,0 )$
or $rR\{ \cos ( \theta +\phi )+i\sin ( \theta +\phi ) \} =( 1,0 )$
or $rR\cos ( \theta +\phi )=1 \text{ and } rR\sin ( \theta +\phi )=0$
or $R=\frac{1}{r},\phi =-\theta$
Thus,
$z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )$

Exponenrial Form

Euler’s formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. It states that for any real number x:
$e^{i \theta}=\cos \theta+i\sin \theta$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions.

When x = π, Euler’s formula evaluates to $e^{i\pi}+1=0$ , which is known as Euler’s identity.

Given Euler’s exponential form,
$e^{i\theta}=\cos \theta +i\sin \theta$
Thus, complex number $z=r( \cos \theta +i\sin \theta )$ is defined as
$z=re^{i\theta}$
The significance of exponential form of complex number is that we can easily compute conjugate and inverse.
For example,
$\bar{z}=re^{- i\theta}$ and $z^{-1} =\frac{1}{r} e^{- i\theta}$

Proof of the Formula

Function Method

Consider the function f(θ) given by
$f(\theta )=\frac {\cos \theta +i\sin \theta }{e^{i\theta }}$
or $f(\theta )= e^{- i\theta } ( \cos \theta +i\sin \theta)$
Differentiating gives by the product rule, we have
$f' (\theta )=0$
Thus, f(θ) is a constant.
Since f(0) = 1, then f(θ) = 1 for all θ, and thus
$1= e^{- i\theta } ( \cos \theta +i\sin \theta)$
or $e^{ i\theta }= \cos \theta +i\sin \theta$
This completes the proof.
Series Method

We know that
$e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$
$\cos x= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}-$
$\sin x= x-\frac{x^3}{3!} +\frac{x^5}{5!}-...$
Using power series expansion, we get
$e^{ i\theta }= \sum_{n=0}^\infty \frac{(i\theta)^n}{n!}$
or $e^{ i\theta }= 1+ \frac{(i\theta)^1}{1!}+ \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!}+ \frac{(i\theta)^4}{4!}+...$
or $e^{ i\theta }= 1+ i\theta- \frac{\theta^2}{2!} -i \frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+...$
or $e^{ i\theta }= \left ( 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}-...\right ) +i \left (\theta-\frac{\theta^3}{3!} +... \right )$
or $e^{ i\theta } = \cos \theta +i\sin \theta$
This completes the proof.

De-Moivre’s theorem

Let $z=r \cos \theta +i\sin \theta$ be a complex number then $z^n =r^n (\cos n\theta +i\sin n\theta )$ where n is a positive integer.
Proof

Case 1: n=1
Then
$z =r (\cos \theta +i\sin \theta )$
or $z^1 =r^1 (\cos 1.\theta +i\sin 1.\theta )$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=1
Case 2: n=2
$z^2 =z.z$
or $z^2 = r (\cos \theta +i\sin \theta ) \times r (\cos \theta +i\sin \theta )$
or $z^2 = r^2 (\cos \theta +i\sin \theta )(\cos \theta +i\sin \theta )$
or $z^2 = r^2 [\cos \theta (\cos \theta +i\sin \theta )+i\sin \theta (\cos \theta +i\sin \theta )$
or $z^2 = r^2 [\cos \theta \cos \theta +i\cos \theta\sin \theta +i\sin \theta\cos \theta -\cos \theta\sin \theta ]$
or $z^2 = r^2 [(\cos \theta \cos \theta-\sin \theta \sin \theta ) +i(\cos \theta\sin \theta +\sin \theta\cos \theta) ]$
or $z^2 = r^2 [\cos (\theta +\theta) +i \sin (\theta+ \theta)]$
or $z^2 = r^2 [\cos 2\theta +i \sin 2\theta]$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=2
Case 3: We assume the same formula is true for n = k, so we have
$(\cos\theta + i\sin\theta)^k = r^k(\cos(k\theta) + i\sin(k\theta))$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=k
Case 4: Now, we prove for n = k + 1,
$[r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos\theta + i\sin\theta)^k r (\cos\theta + i\sin\theta)$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos(k\theta) + i\sin(k\theta)) r(\cos\theta + i\sin\theta)$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} = r^{k+1}[(\cos(k\theta) \cos\theta-\sin(k\theta)\sin\theta )+i (\cos\theta\sin(k\theta) + \sin\theta\cos(k\theta))]$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k\theta+\theta)+i \sin(k\theta+\theta )]$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k+1)\theta+i \sin(k+1)\theta]$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=k+1
Using case 1-case 4, for any number $n \in Z$ , we have
$[r(\cos\theta + i\sin\theta)]^n =r^n[\cos(n\theta)+i \sin(n\theta)]$

Example 1

Compute $(1+i)^6$
Solution
Since
$1+i= \sqrt{2} \left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right )$
or $1+i= \sqrt{2} \left ( \cos \frac{\pi }{4} +i\sin \frac{\pi }{4} \right )$
We get
$r=\sqrt{2}$ and $\theta=\frac{\pi }{4}$
Thus,
$( 1+i )^6 =\sqrt{2}^6 \left [ \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right ]^6$
or $( 1+i )^6 =\sqrt{2}^6[ \cos \left (6 \frac{\pi }{4} \right )+i\sin \left (6 \frac{\pi }{4} \right )]$
or $( 1+i )^6 =8 (\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})$
or $( 1+i )^6 =-8i$

nth root of Complex number

If $Z=r(\cos \theta +i\sin \theta )$ be a complex number then the nth root of z is
$\sqrt[n]{Z}=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )$
Proof
Given that Z is a complex number. Also let, nth root of Z is W such that $W=R(\cos \phi +i\sin \phi )$
Now we have
$\sqrt[n]{Z}=W$
or $W^n=Z$
or $[R(\cos \phi +i\sin \phi )]^n=r(\cos \theta +i\sin \theta )$
or $R^n(\cos (n\phi) +i\sin (n\phi) )=r(\cos \theta +i\sin \theta )$
Equating real and Imaginary parts, we get
$R^n=r$ and $\cos (n\phi)= \cos \theta$ and $\sin (n\phi) =\sin \theta$
or $R=\sqrt[n]{r}$ and $n\phi= \theta +2k \pi$
or $R=\sqrt[n]{r}$ and $\phi= \frac{(\theta +2k\pi )}{n}$
Thus, nth root of $Z=r(\cos \theta +i\sin \theta )$ is
$W=R(\cos \phi +i\sin \phi )$
or $W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )$

square roots of i

Find the square roots of i

We know that
$\sqrt[n]{z}=\sqrt[n]{r} \left [ \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ]$

Solution
Since $i=\cos \frac{\pi }{2} +i\sin \frac{\pi }{2}$ , we get
r=1 and $\theta=\frac{\pi }{2}$
Hence the first square roots of i is
$w_1[i=0] =\cos \frac{\frac{\pi }{2}+0}{2} +i\sin \frac{\frac{\pi }{2}+0}{2} =\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$
The second square root of i is

$w_2[i=1] =\cos \frac{\frac{\pi }{2}+2\pi }{2} +i\sin \frac{\frac{\pi }{2}+2\pi }{2} =\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$

Conic Section

Posted on January 3, 2024January 7, 2024 by MEAN

[latexpaage]

Conic Section

औपचारिक रूपमा, ग्रीकमा लगभग 500 to 200 BC को अवधिमा Conic sections पत्ता लगाइएको मानिन्छ। त्यतिखेर अपोलोनिस (200BC)ले Conic sections को बारेमा चर्चा गरेको पाईन्छ। यद्पि, सत्रौं शताब्दीको सुरुबाट मात्र Conic sections को व्यापक प्रयोग भएको पाईन्छ। आजका दिनहरूमा, प्रकृतिमा हुने धेरै प्रक्रियाहरूलाई मोडेल गर्न conic sections महत्त्वपूर्ण छन्। उदाहरण को लागी, universal bodies को locus कोनिक्स (सर्कल, अण्डाकार, प्याराबोला, र हाइपरबोला) हो ।

The term “conic section” refers to the geometric shapes formed by the intersection of a plane with a cone. The cone is not necessarily a right circular cone. There are many conic sections on the basis of the angle of cutting. In every cases, conic section is the section of cone by a plane.

The ancient Greek mathematician Apollonius, also known as “Great Geometer,” made significant contributions to the understanding of conic sections around 200 BC. Apollonius explored the properties of ellipses, parabolas, and hyperbolas, categorizing them based on their unique characteristics.

Cone

In mathematics, cone is defined a three-dimensional surface
traced out by a straight line
passing through a fixed point and
moving around a fixed line.
In this definition of cone,
The straight line is called generator
The fixed point is called vertex
The fixed line is called axis

Conic भनेको डबल शंकुको cone लाई एउटा सतहले काट्दा बन्ने वक्र रेखा हो। सामान्यतया यसलाई right circular cone मा हेर्ने गरिन्छ, तर यो जुनसुकै cone मा पनि परिभाषित हुन्छ ।

Conic Section: The Geometry

Cone लाई एउटा plane ले काट्दा बन्ने plane curve (cross section) लाई conic section भनिन्छ ।

Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
Based on this intersection, there are seven types of conic section.
These seven types of conic section are given below

	Conic	Vertex	Generator	Axis
1	Point	Cuts	outside
2	Line	Cuts	touches
3	Line Pair	Cuts	inside
4	Circle	Misses		Right angle
5	Ellipse	Misses		Not right angle
6	Parabola	Misses	Parallel to
7	Hyperbola	Misses	Not Parallel to

According to this table, for example,
Parabola: a parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which

the plane misses the vertex and
the plane is parallel to the generator.

Hyperbola: Hyperbole is a conic section obtained by section of a cone by a plane in which

the plane misses the vertex of cone
the plane is not parallel to the generator of cone

Similarly, we can define other conic sections

Notice that, from the table above we see that in some intersection, plane does not pass through the vertex of the cone. When the plane does pass through the vertex, the resulting conic is a degenerate conic section.

Conic Section: The Algebra

In analytic geometry, conic section can be defined in algebraic expression. This algebraic forms of conic section is called analytic representation.

In analytic geometry,
Conic section can be defined based on the definition of circle.

Circle

Circle is defined a locus of point whose
distance from a fixed point = constant

In this definition of circle,

the constant distance is called radius.
the fixed point is called center

Based on this definition of circle, we can define conic section.

Conic Section

Conic section is defined a locus of a point whose
$ \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} $= constant
In this definition of conic section,

the constant ratio is called eccentricity, it is denoted by e.
the fixed point is called focus.
the fixed line is called directrix.

Classification of Conic Section

Based on the value of e, conic section can be classified into three standard types. These three standard types are

Parabola (e =1)
Ellipse (e <1)
Hyperbola (e >1)

Circle (e =0)
Straight Line (e =∞)

General equation of Conic

If the general equation of second degree $ ax^2 +2hxy + by^2 +2gx + 2fy + c = 0$ represents a pair of striaght lines then the discriminat must be perfect square, thus
$(2hy+2g)^2-4(a)(by^2+2fy+c)$ is perfect square
or $(hy+g)^2-a(by^2+2fy+c)$ is perfect square
or $(h^2-ab)y^2+2(hg-af)y +(g^2-ac)$ is perfect square
Again, $(h^2-ab)y^2+2(hg-af)y +(g^2-ac)$ is perfect square if its discriminant is zero
Thus,
$ 4(hg-af)^2-4(h^2-ab)(g^2-ac)=0$
or $ (hg-af)^2-(h^2-ab)(g^2-ac)=0$
or $ abc+2fgh-af^2-bg^2-ch^2=0$

Therefore, the discriminant is
$ \Delta= abc+2fgh-af^2-bg^2-ch^2 $

$ Δ = 0, h^2=ab$ then the conic is pair of straight lines
$ Δ = 0, \frac{a}{h}= \frac{h}{b}=\frac{g}{f} $ then the conic is parallel lines
$ Δ \ne 0, a=b\ne 0, h=0$ then the conic is circle
$ Δ \ne 0, h^2 < ab$ then the conic is ellipse
$ Δ \ne 0, h^2=ab$ then the conic is parabola
$ \Delta \ne 0, h^2 >ab$ then the conic is hyperbola
$ Δ \ne 0, h^2 >ab, a+b=0$ then the conic is rectangular hyperbola

The equation of the conic whose center is at the origin is of the form
$ ax^2+by^2+2hxy+1=0 $
This conic is called central conic

Parabola

Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
So,
Parabola caan be defined as follows.

Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which
the plane misses the vertex and
the plane is parallel to the generator

Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which if
α= angle between generator and axis and
β= angle between plane and axis,
and
α=β
then
the section is a parabola,
in which
eccentricity = $\frac{\cos \beta}{\cos \alpha}$
here
the eccentricity is the measure of how far the conic deviates from being circular
Parabola is a plane curve defined a locus of a point in which
$ \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} $= constant (e=1)

In this definition of conic section,
the constant ratio is called eccentricity, it is denoted by e.
the fixed point is called focus.
the fixed line is called directrix.

Parabola is a plane curve defined a locus of a point in which the distance from a fixed point (or focus) and distance from a fixed line (or directrix) is always equal.
Parabola is a plane curve defined a locus of a point which is always equidistant from a fixed point (or focus) to a fixed line (or directrix)

Focus: The fixed point of parabola is called focus
Directrix: The fixed line of parabola is called directrix
Axis: The straight line passing through focus and perpendicular to directrix is called axis
Vertex: The meeting point of axis and parabola is called vertex
Latus rectum: the chord passing through focus and perpendicular to axis is called latus rectum.
The distance between the meeting points of latus rectum to the parabola is called length of latus rectum.

Equation of Parabola

Let C be a parabola whose
Focus is F (a,0)
Directrix is $l: x = -a$
Vertex is O: (0,0)
Take any point P(x,y) on parabola C,

Draw PA ⊥ $l$ then A (-a,y)
Join F and P

By the definition of parabola
PA = PF
or $ (x+a)^2=(x-a)^2+y^2 $
or $x^2+2ax+a^2=x^2-2ax+a^2+y^2$
or $2ax=-2ax+y^2 $
or $y^2=4ax $

Summary on Equation of Parabola

The basic parameters of parabola are summarized as below

Parabola	Parabola	Parabola	Parabola	Parabola
Equation	$ y^2 =4 a x $	$ x^2 =4 a y $	$ (y-k)^2 =4 a (x-h) $	$ (x-h)^2 =4 a (y-k) $
Vertex	(0,0)	(0,0)	(h,k)	(h,k)
Focus	(a,0)	(0,a)	(h+a,k)	(h,k+a)
Directrix	x=-a	y=-a	x=h-a	y=k-a
Axis	y=0	x=0	y=k	x=h
Axis of Symmetry	x-axis	y-axis	y=k	x=h
Endpoints of Latus Rectum	(a,±2a)	(±2a,a)	(h+a,k±2a)	(h±2a,k+a)

Ellipse

Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
So,
Ellipse can be defined as follows.

Ellipse is conic section defined as a plane curve obtained by intersection of a cone and a plane in which
the plane misses the vertex and
the plane is NOT at right angle with the axis

Ellipse is conic section defined as a plane curve obtained by intersection of a cone and a plane in which if
α= angle between generator and axis and
β= angle between plane and axis,
and
α < β < 90
then
the section is a ellipse,
in which
eccentricity = $\frac{\cos \beta}{\cos \alpha}$
here
the eccentricity is the measure of how far the conic deviates from being circular
Ellipse is a plane curve defined a locus of a point in which
$ \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} $= constant (e <1)

In this definition of conic section,
the constant ratio is called eccentricity, it is denoted by e.
the fixed point is called focus.
the fixed line is called directrix.

Ellipse is a plane curve defined a locus of a point in which the distance from a fixed point (or focus) is always less than the distance from a fixed line (or directrix)
Ellipse is a plane curve defined a locus of a point whose sum of distances from two fixed points (foci) is always a constant.

In this definition

Foci: The two fixed points of ellipse
Directrix: The two fixed line of ellipse
Axis: The straight line passing through focus and perpendicular to directrix
Major axis: The straight line passing through the foci
Minor axis: The straight line passing through the center and perpendicular to the major axis
Length of major axis: The distance between the vertices on major axis
Length of minor axis: The distance between the meeting points of minor axis with ellipse
Vertices: The meeting points of the major axis with ellipse
Centre of ellipse: The middle point of the join of foci
Latus rectum: The chord passing through focus and perpendicular to major axis

Please Note that

Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis.
Each endpoint of the major axis is the vertex of the ellipse, and each endpoint of the minor axis is a co-vertex of the ellipse.
The center of an ellipse is the midpoint of both the major and minor axes.
The axes are perpendicular at the center.
The foci always lie on the major axis

There are four variations of the standard form of the ellipse. These variations are categorized first by
the location of the center (the origin or not the origin),
and by
the position (horizontal or vertical).

Ellipse: Proof of Basic Facts

c=ae
If Z and Z’ are the directrix, F(c,0) and F'(-c,0) are the foci, and A(a,0) and A;(-a,0) are the vertices of ellipse, then by the definition of conic we have
$\frac{A’F}{A’Z}=e$
or $A’F=A’Z e$ (1)
Similarly, we have
$\frac{AF}{AZ}=e$
or $AF=AZ e$ (2)
Substracting (2) from (1), we get
$A’F-AF=(A’Z-AZ) e$
or$(A’O+OF)-(AO-OF)=(A’A) e$
or$2OF=2a e$
or$2c=2a e$
or$c=a e$

sum of distance from Foci is 2a.
the distance of a point A(a,0) from fous F(c,0) is
a−(c)=a-c.
the distance of a point A(a,0) from fous F'(-c,0) is
a−(-c)=a+c
The sum of the distances from Foci is
(a+c)+(a−c)=2a
Relation between a, b, c.
$2 \sqrt{b^2+c^2}=2a$
or$b^2+c^2=a^2$

Equation of Ellipse

Let C be an ellipse whose
foci are (−c,0) and (c,0).
center is O: (0,0)
Take any point P(x,y) on ellipse C,

If (a,0) is a vertex of the ellipse, then the distance from (−c,0) to (a,0) is
a−(−c)=a+c.
The distance from (c,0) to (a,0) is
a−c.
The sum of the distances from the foci to the vertex is
(a+c)+(a−c)=2a
If (x,y) is a point on the ellipse, then we can define the following variables:
d1=the distance from (−c,0)to (x,y)
d2=the distance from (c,0)to (x,y)
By the definition of an ellipse,
d1+d2=2a

Here
$d_1+d_2=2a$
or $\sqrt{(x+c)^2+y^2}+\sqrt{(x−c)^2+y^2}=2a$
or $\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x−c)^2+y^2}$
or $(x+c)^2+y^2=\left [2a-\sqrt{(x−c)^2+y^2} \right] ^2$
or $x^2+2xc+c^2+y^2=4a^2-4a\sqrt{(x−c)^2+y^2} +(x−c)^2+y^2 $
or $x^2+2xc+c^2+y^2=4a^2-4a\sqrt{(x−c)^2+y^2} +x^2-2xc+c^2+y^2 $
or $2xc=4a^2-4a\sqrt{(x−c)^2+y^2} -2xc$
or $4xc-4a^2=-4a\sqrt{(x−c)^2+y^2}$
or $xc-a^2=-a\sqrt{(x−c)^2+y^2}$
or $\left[xc-a^2\right]^2= \left[ -a\sqrt{(x−c)^2+y^2} \right ]^2$
or $c^2x^2-2a^2cx+a^4= a^2 \left[(x−c)^2+y^2 \right ]$
or $c^2x^2-2a^2cx+a^4= a^2 (x^2-2xc+c^2+y^2 ) $
or $c^2x^2-2a^2cx+a^4= a^2x^2-a^22xc+a^2c^2+a^2y^2 $
or $a^2x^2-c^2x^2+a^2y^2=a^4-a^2c^2 $
or $x^2(a^2-c^2)+a^2y^2=a^2(a^2-c^2) $
or $x^2b^2+a^2y^2=a^2b^2 $ $a^2-c^2=b^2$
or $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 $

Thus, the standard equation of an ellipse is
$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 $
This equation defines an ellipse centered at the origin.

If a>b, the ellipse is stretched further in the horizontal direction
if b>a the ellipse is stretched further in the vertical direction.
When c = 0, both foci merge together with the center of the ellipse and so the ellipse becomes a circle
When c = a, then b = 0. The ellipse reduces to the line segment joining the two foci

Summary of parameters in an Ellipse

	Ellipse	Ellipse	Ellipse	Ellipse
Equation	$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$; a > b	$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$; a < b	$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$; a > b	$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$; a < b
Center	(0,0)	(0,0)	(h,k)	(h,k)
Vertex	$ (\pm a,0) $	$ (0,\pm b) $	$ (h\pm a,k) $	$ (h,k\pm b) $
Focus	$ (\pm ae,0) $	$ (0,\pm be) $	$ (h\pm ae,k) $	$ (h,k\pm be) $
Directrix	$ x=\pm \frac{a}{e} $	$ y=\pm \frac{b}{e} $	$ x=h\pm \frac{a}{e} $	$ y=k\pm \frac{b}{e} $
Length Rectum	$ 2 \frac{b^2}{a} $	$ 2 \frac{a^2}{b}$	$ 2 \frac{b^2}{a} $	$ 2 \frac{a^2}{b}$
Eccentricity	$ b^2=a^2(1-e^2) $	$ a^2=b^2(1-e^2) $	$ b^2=a^2(1-e^2) $	$ a^2=b^2(1-e^2) $

TSC Math Online Class

Posted on December 28, 2023December 28, 2023 by MEAN

NOTICE!TSC Math Mavi Second Paper, online class लिन ईच्छुक साथीहरुको लागी अवसर!!!

कक्षा शुरु: २०८० पौष १५ बाट
This class is only for the second paper (Mathematics)
The class will be provided for 100 days (50 Days Section A, 50 Days Section B)

Class Model: (online class with Zoom) Special features!1. Recorded Video Available (200+)2. Live Class Video Available (100+)3. Complete and Interactive Digital Book Available (All Chapters) 4. Complete PDF note Available (All Chapters) 5. Sample Questions and Answers Available (20 Questions and answers, two from each chapter)
Contact
9847360402
Mathematics Education Advocacy Nepal (MEAN) [website: https://mean.edu.np/]

If you are interested, please fill up the form

Unit Circle

Posted on December 27, 2023December 27, 2023 by MEAN

Unit circle

Radius को मान 1 भएको, केन्द्र (0,0) भएको circle लाई unit circle भनिन्छ। Trigonometry मा unit circle को ज्यादै महत्व छ। यसैबाट trigonometric ratio, quadrant सम्बन्धि नियम, trigonometric identity पत्ता लगाउन सकिन्छ ।

कोण (0 देखि 360 डिग्रि सम्मको) दिएमा trigonometry का छ वटा standard ratio हरु sin, cos र tan साथै cosec, sec, cot को मान पत्ता लगाउन सकिन्छ ।
trigonometry का छ वटा standard ratio हरुको मान दिएमा त्यसको corresponding कोण पत्ता लगाउन सकिन्छ.
trigonometry सम्बन्धी Quotient, Reciprocal, Co-function, Even-odd, Pythagorean जस्ता identities प्रमाणीत गर्न सकीन्छ ।

Sinθ र cosθ को मान

केन्द्र O को मान O(0,0) भएको, radius (r=OA) को मान 1 भएको circle C मा एउटा moving point (arbitrary point) P(x,y) लिइएको छ, जसले circle को केन्द्र O मा धनात्मक दिशातिर θ डिग्रिको कोण बनाएको छ।
∡AOP=θ
यहा,
the x-coordinate of P(x,y) on the unit circle gives the cosine value of the angle θ. The y-coordinate gives the sine value of the angle θ
OP=r=1
OQ=x=cosθ
PQ=y=sinθ
तसर्थ,
$\sin^2 \theta+\cos^2\theta=1$

बिन्दु P(x,y) लाई चलाउनुहोस र sinθ वा cosθ को मान पत्ता लगाउनुहोस

tanθ र secθ को मान

केन्द्र O को मान O(0,0) भएको, radius (r=OA) को मान 1 भएको circle C मा बिन्दु A बाट circle C मा tangent खिचिएको छ । यस tangent रेखामा एउटा moving point (arbitrary point) P(x,y) लिइएको छ, जसले circle को केन्द्र O मा धनात्मक दिशातिर θ डिग्रिको कोण बनाएको छ।
∡AOP=θ
यहा, त्रिभुज OAP मा,
the segment AP on the triangle OAP gives the tangent value of the angle θ. The segment OP on the triangle OAP gives the secant value of the angle θ
OA=r=1
AP=tanθ
OP=secθ
तसर्थ,
$\sec^2 \theta-\tan^2\theta=1$

बिन्दु P(x,y) लाई चलाउनुहोस र tanθ वा secθ को मान पत्ता लगाउनुहोस

cotθ र cosecθ को मान

केन्द्र O को मान O(0,0) भएको, radius (r=OB) को मान 1 भएको circle C मा बिन्दु B बाट circle C मा tangent खिचिएको छ । यस tangent रेखामा एउटा moving point (arbitrary point) P(x,y) लिइएको छ, जसले circle को केन्द्र O मा धनात्मक दिशातिर θ डिग्रिको कोण बनाएको छ।
∡AOP=θ
यहा, त्रिभुज OBP मा,
the segment BP on the triangle OBP gives the cotangent value of the angle θ. The segment OP on the triangle OBP gives the cosecant value of the angle θ
OB=r=1
BP=cotθ
OP=cosecθ
तसर्थ,
$\csc^2 \theta-\cot^2\theta=1$

बिन्दु P(x,y) लाई चलाउनुहोस र cotθ वा cosecθ को मान पत्ता लगाउनुहोस

The details of Trigonometric ratio and its name

“sine” भन्ने शब्द Latin भाषाको “sinus” भन्ने शब्दबाट ल्याईएको हो जसको अर्थ “bend” or “curve” भन्ने हुन्छ, जसले ग्राफमा sinusoidal or S-wave-like curve बनाउदछ ।
“cosine” भन्ने शब्द Latin भाषाको “cosinus” भन्ने शब्दबाट ल्याईएको हो जसको अर्थ “sine of the complementary angle” भन्ने हुन्छ, जसले ग्राफमा C-wave-like curve बनाउदछ ।
“tangent” भन्ने शब्द Latin भाषाको “tangens” भन्ने शब्दबाट ल्याईएको हो जसको अर्थ “touching” or “to touch” भन्ने हुन्छ, जसले ग्राफमा length of the line segment on tangent (toward-x-axis) लाई जनाउदछ ।
“secant” भन्ने शब्द Latin भाषाको “secare” भन्ने शब्दबाट ल्याईएको हो जसको अर्थ “to cut” or “cutting” भन्ने हुन्छ, जसले unit circle मा बन्ने right-angled triangle मा हुने tangent segment लाई cut गर्ने hypoteneous segment लाई जनाउदछ । (toward x-axis)
“cotangent” भन्ने शब्द Latin भाषाको “cotangens” भन्ने शब्दबाट ल्याईएको हो जँहा “co” को अर्थ “complement” भन्ने हुन्छ, जसले tangent को complement angle लाई जनाउदछ ।
“cosecant” भन्ने शब्द Latin भाषाको “cosecans” भन्ने शब्दबाट ल्याईएको हो जसको अर्थ “the one who cuts” or “the cutter” भन्ने हुन्छ, जसले unit circle मा बन्ने right-angled triangle मा हुने cotangent segment लाई cut गर्ने hypoteneous segment लाई जनाउदछ । (toward y-axis)

Visual Proof (Trigonometry)

Posted on December 26, 2023December 27, 2023 by MEAN

Visual Proof of sin(α+β) and cos(α+β)

$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
$\cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$

केन्द्र O को मान O(0,0) भएको, radius (r=OB) को मान 1 भएको circle C मा रेखा OA ले केन्द्र O मा धनात्मक दिशातिर α कोण र रेखा OB ले केन्द्र O मा धनात्मक दिशातिर β कोण बनाउदछ। बिन्दु B बाट रेखा OA मा लम्ब BA⊥OA खिचिएको छ । अब, बिन्दु O, A र B बाट बन्ने आयात OPQR बनाईएको छ जसमा
Right angled triangle ⊿OAB बाट,
∡AOB=β
OB=1
OA=cosβ
AB=sinβ
Right angled triangle ⊿OAP बाट,
∡AOP=α
OP=cosαcosβ
AP=sinαcosβ
Right angled triangle ⊿AQB बाट,
∡QAB=α
QA=cosαsinβ
QB=sinαsinβ
Right angled triangle ⊿OBR बाट,
∡OBR=α+β
OB=1
BR=cos(α+β)
OR=sin(α+β)
We know that
OR=PA+AQ
or sin(α+β)=sinαcosβ+cosαsinβ
Also, we know that
BR=OP-QB
or cos(α+β)=cosαcosβ-sinαsinβ

Visual Proof of tan(α+β)

$\tan(\alpha+\beta)= \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$

केन्द्र O को मान O(0,0) भएको, radius (r=OP) को मान 1 भएको circle C मा रेखा OA ले केन्द्र O मा धनात्मक दिशातिर α कोण र रेखा OB ले केन्द्र O मा धनात्मक दिशातिर β कोण बनाउदछ। बिन्दु A बाट रेखा OB मा लम्ब BA⊥OA खिचिएको छ । अब, बिन्दु O, A र B बाट बन्ने आयात OPQR बनाईएको छ जसमा
Right angled triangle ⊿OAP बाट,
∡AOP=α
OP=1
AP=tanα
OA=secα
Right angled triangle ⊿OAB बाट,
∡AOB=β
OB=tanβsecα
Right angled triangle ⊿AQB बाट,
∡QAB=α
QA=tanβ
QB=tanαtanβ
Right angled triangle ⊿OBR बाट,
∡OBR=α+β
We know that
BR=OP-QB
Therefore
$\tan(\alpha+\beta)= \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$

Visual Proof of cot(α+β)

$\cot(\alpha+\beta)= \frac{\cot \alpha \cot \beta-1}{\cot \alpha + \cot \beta}$

केन्द्र Q को मान O(0,0) भएको, radius (r=QB) को मान 1 भएको circle C मा रेखा OA ले बिन्दु O मा धनात्मक दिशातिर α कोण र रेखा OB ले बिन्दु O मा धनात्मक दिशातिर β कोण बनाउदछ। बिन्दु A बाट रेखा OB मा लम्ब BA⊥OA खिचिएको छ । अब, बिन्दु O, A र B बाट बन्ने आयात OPQR बनाईएको छ जसमा
Right angled triangle ⊿AQB बाट,
∡QAB=α
QA=1
AB=cosecα
QB=cotα
Right angled triangle ⊿OAB बाट,
∡AOB=β
OA=cosecαcotβ
Right angled triangle ⊿OAP बाट,
∡AOP=α
OP=cotαcotβ
AP=cotβ
Right angled triangle ⊿OBR बाट,
∡OBR=α+β
We know that
BR=OP-QB=cotαcotβ-1
Therefore
$\cot(\alpha+\beta)= \frac{\cot \alpha \cot \beta-1}{\cot \alpha + \cot \beta}$

Reflection

Posted on December 19, 2023December 27, 2023 by MEAN

ज्यामितिमा स्थानान्तरण लामो समयदेखि व्यापक चासोको विषय बनेको थियो। युक्लिडको आधिकारिक वा स्वयंसिद्ध सिद्धान्त नभएपनि सामान्यतया स्थानान्तरणलाई सुपरपोजिसनको सिद्धान्त (उठाउर अर्कोमा राख्नु) मा आधारित गरिएको थियो । जब १९ औं शताब्दीमा (1872 मा) फ्लेक्स क्लेनले – एर्लान्जेन विश्वविद्यालयमा – ज्यामितिमा एक नयाँ परिप्रेक्ष्य प्रस्तावित गरे तबदेखी स्थानान्तरणलाई एर्लान्जेन कार्यक्रमको रूपमा चिनिन थालिएको छ।

Transformation Geometry

Transformation Geometry is a branch of mathematics that utilize analytic geometry and algebraic function to study geometric invariants. It was introduced (developed) by Flex Klein in19th century through Erlangen programme. According to him, theer are five Kinds of Transformations:

Rigid motions: Plane geometry of congruent figures, preserve distances (e.g., Euclidean transformations)
Similarity transformations: The familiar geometry with similar figures.
Similarities preserve angles and ratios between distances (e.g., resizing)
Affine (matrix) transformations: Geometry of computer animation. Rectangles and parallelograms the same
Affine transformations preserve parallelism (e.g., scaling, shear)
Projective Geeometry: Geometry of photo
Projective transformations preserve collinearity
Continuous (topological) transformations: Any loop is a circle. Any path is a segment

In transformation geometry, the idea of flips, slides, and turn are key features, which will be analyzed based on its mapping. This mapping are basically of two kinds.

Isometric mapping (transformation)
Non-isometric mapping (transformation)

Geometric transformation involves both analytic and algebraic geometry because it can be approached using the graphical perceptive, and algebraic computational theory.

Reflection

Reflection is very common topic in physics and mathematics. When an object is place before the mirror, the image is formed behind the mirror. This is called reflection.

Definition: Reflection

Let l be a line and A be a point.
Then, reflection about the line l is denoted by $T_1$ , which is a transformation on a plane, where

if A ∉l , then T_l (A) = A’
if A ∈l , then T_l(A)=A

such that

the mispoint of AA’ lies on l
the segment AA’is perpendiculat to l

We start a short discussion for reflection in mathematics as below.

Let us take a geometrical figures like triangle. Now, we hold the triangle vertices, one-by-one, about a line source $l$ in a grid. Then look at the image (shadow) of the triangle on the grid. The shadow have the same size as the original triangle has, but in opposite direction.

In the above figure, ∆XYZ is a triangle and its shadow is ∆X’Y’Z’.
By actual measurements it can be seen that
∡X = ∡X’, ∡Y = ∡Y’, ∡Z = ∡Z
XY=X’Y’, YZ=Y’Z’, XZ=X’Z’
Thus, the two triangles are congruent
The important note is that

l ⊥ bisector of XX’
l ⊥ bisector of YY’
l ⊥ bisector of ZZ’

Here $l$ is said to be line of reflection

In summary,

The image is found opposite to the mirror line $l$ .
Mirror line $l$ is the perpendicular bisector of the line joining the object and its image.
X’ is called the reflection (image) of X and so on
$l$ is called the axis of reflection.

Thus, reflection can be seen in water, in a mirror, in glass. In mathematics, an object and its reflection have the same shape and size, but the figures face in opposite directions . Also, a reflection needs

Object
Line of Reflection

Quadratic Equation [BCB Ex6.1]

Posted on August 16, 2023December 27, 2023 by MEAN

Determine the nature of roots of each of the following equations
1. $x^2-6x+5=0$
  
  Solution 👉 Click Here
  
  Solution :1a
2. $x^2-4x-3=0$
  
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  Solution :1b
3. $x^2-6x+9=0$
  
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  Solution :1c
4. $4x^2-4x+1=0$
  
  Solution 👉 Click Here
  
  Solution :1d
5. $2x^2-9x+35=0$
  
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  Solution :1e
6. $4x^2+8x-5=0$
  
  Solution 👉 Click Here
  
  Solution :1f
For what values of p will the equation $5x^2-px+45=0$ ?

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Solution :2
if the equation $x^2+2(k+2)x+9k=0$ has equal roots, find k.

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Solution :3
For what value of a will the equation $x^2-(3a-1)x+2(a^2-1)=0$ have equal roots?

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Solution :4
If the roots of the equation $(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0$ are equal, then show that $\frac{a}{b}=\frac{c}{d}$

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Solution :5
Show that the roots of the equation $(a^2-bc)x^2+2(b^2-ca)x+(c^2-ab)=0$ will be equal, if either $b=0$ or $a^3+b^3+c^3-3abc=0$

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Solution :6
If $a,b,c$ are rational and $a+b+c=0$ , show that the roots $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ are rational.

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Solution :7
Prove that the roots of the equation $(x-a)(x-b)=k^2$ are real for all values of k.

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Solution :8
Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.

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Solution :9
If the roots of the quadratic equation $qx^2+2px+2q=0$ are real and unequal, prove that the roots of the equation $(p+q)x^2+2qx+(p-q)=0$ are imaginary

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Solution :10

	Ellipse	Ellipse	Ellipse	Ellipse
Equation	\( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\); a > b	\( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\); a < b	\( \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\); a > b	\( \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\); a < b
Center	(0,0)	(0,0)	(h,k)	(h,k)
Vertex	\( (\pm a,0) \)	\( (0,\pm b) \)	\( (h\pm a,k) \)	\( (h,k\pm b) \)
Focus	\( (\pm ae,0) \)	\( (0,\pm be) \)	\( (h\pm ae,k) \)	\( (h,k\pm be) \)
Directrix	\( x=\pm \frac{a}{e} \)	\( y=\pm \frac{b}{e} \)	\( x=h\pm \frac{a}{e} \)	\( y=k\pm \frac{b}{e} \)
Length Rectum	\( 2 \frac{b^2}{a} \)	\( 2 \frac{a^2}{b}\)	\( 2 \frac{b^2}{a} \)	\( 2 \frac{a^2}{b}\)
Eccentricity	\( b^2=a^2(1-e^2) \)	\( a^2=b^2(1-e^2) \)	\( b^2=a^2(1-e^2) \)	\( a^2=b^2(1-e^2) \)