Math 11 - 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

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.

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

Solution 👉 Click Here

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

Solution 👉 Click Here

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

Solution 👉 Click Here

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 $\frac{a}{b}=\frac{c}{d}$

Solution 👉 Click Here

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 $a^3+b^3+c^3-3abc=0$

Solution 👉 Click Here

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.

Solution 👉 Click Here

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

Solution 👉 Click Here

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

Solution 👉 Click Here

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

Solution 👉 Click Here

Solution :10

Quadratic Equation [BCB Ex6.2]

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

From the equation whose roots are
1. 3,-2
  
  Solution 👉 Click Here
  
  Solution :1a
2. -5,4
  
  Solution 👉 Click Here
  
  Solution :1b
3. $\sqrt{3},-\sqrt{3}$
  
  Solution 👉 Click Here
  
  Solution :1c
4. $\frac{1}{2} (-1+\sqrt{5}),\frac{1}{2} (-1-\sqrt{5})$
  
  Solution 👉 Click Here
  
  Solution :1d
5. -3+5i,-i-5i
  
  Solution 👉 Click Here
  
  Solution :1e
6. a+ib,a-ib
  
  Solution 👉 Click Here
  
  Solution :1f
1. Find a quadratic equation whose roots are twice the roots of $4x^2+8x-5=0$
  
  Solution 👉 Click Here
  
  Solution :2a
2. Find a quadratic equation whose roots are reciprocals of the roots of $3x^2-5x-2=0$
  
  Solution 👉 Click Here
  
  Solution :2(b)
3. Find a quadratic equation whose roots are greater by h than the roots of $x^2-px+q=0$
  
  Solution 👉 Click Here
  
  Solution :2(c)
4. Find a quadratic equation whose roots are the squares of the roots of $3x^2-5x-2=0$
  
  Solution 👉 Click Here
  
  Solution :2(d)
Find a quadratic equation with rational coefficients one of whose roots is
1. 4+3i
  
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  Solution :3a
2. $\frac{1}{5+3i}$
  
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  Solution :3b
3. $2+\sqrt{3}$
  
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  Solution :3c
Find the value of k so that the equation
1. $2x^2+kx-15=0$ has one root 3
  
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  Solution :4a
2. $3x^2+kx-2=0$ has roots whose sum is equal to 6
  
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  Solution :4b
3. $2x^2+(4-k)x-17=0$ has roots equal but opposite in sign
  
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  Solution :4c
4. $3x^2+(5+k)x+8=0$ has roots numerically equal but opposite in sign
  
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  Solution :4d
5. $3x^2+7x+6-k=0$ has one root equal to zero
  
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  Solution :4e
6. $4x^2-17x+k=0$ has the reciprocal roots
  
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  Solution :4f
7. $4x^2+kx+5=0$ has roots whose difference is $\frac{1}{4}$
  
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  Solution :4g
Show that -1 is a root of the equation $a+b-2c$ x^2+(2a-b-c)x+(c+a-2b)=0\). Find the other root.

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Solution :5
Find the value of m for which the equation $(m+1)x^2+2(m+3)x+(2m+3)=0$ will have (a) reciprocal roots (b) one root zero.

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Solution :6
If the roots of the equation $x^2+ax+c=0$ differ by 1, prove that $a^2=4c+1$

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Solution :7
If are the roots of the equation , find the equation whose roots are
1. $\alpha ^2 \beta ^{-1}$ and $\beta ^2 \alpha ^{-1}$
  
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  Solution :8a
2. $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$
  
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  Solution :8(b)
If are the roots of the equation , find the equation whose roots are
1. $\alpha ^2 \beta ^{-1}$ and $\beta ^2 \alpha ^{-1}$
  
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  Solution :9a
2. $\alpha ^3$ and $\beta ^3$
  
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  Solution :9b
3. $(\alpha-\beta)^2$ and $(\alpha+\beta)^2$
  
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  Solution :9c
4. the reciprocal of the roots of given equation
  
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  Solution :9d
1. If the roots of the equation $ax^2+bx+c=0$ be in the ratio of 3:4, prove that $12b^2=49ac$
  
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  Solution :10a
2. If one root of the equation $ax^2+bx+c=0$ be four times the other root, show that $4b^2=25ac$
  
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  Solution :10b
3. For what values of m, the equation $x^2-mx+m+1=0$ may have its root in the ratio 2:3
  
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  Solution :10c
1. If $\alpha, \beta$ are the roots of the equation $px^2+qx+q=0$ , prove that $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0$
  
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  Solution :11a
2. If roots of the equation $lx^2+nx+n=0$ be in the ratio of p:q, prove that $\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$
  
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  Solution :11b
If one root of the equation $ax^2+bx+c=0$ be square of the other root, prove that $b^3+a^2c+ac^2=3abc$

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Solution :12

Quadratic Equation[BCB Ex6.3]

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

Show that each pair of following equations has a common root
1. $x^2-8x+15=0$ and $2x^2-x-15 =0$
  
  Solution 👉 Click Here
  
  Solution :1(a)
  Given quadratic equations are
  $x^2-8x+15=0$ and $2x^2-x-15 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=1,b_1=-8,c_1=15,a_2=2,b_2=-1,c_2=-15$
  We know that, condition for two quadratic equations having a common root is
  $(a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[2.(15)-1(-15)]^2 = [1(-1)-2(-8)] [(-8)(-15)-(-1)(15)]$
  or $[30+15]^2 = [-1+16] [120+15]$
  or $2035 =2035$
  Thus, $x^2-8x+15=0$ and $2x^2-x-15 =0$ has a common root.
2. $3x^2-8x+4=0$ and $4x^2-7x-2 =0$
  
  Solution 👉 Click Here
  
  Solution :1(b)
  Given quadratic equations are
  $3x^2-8x+4=0$ and $4x^2-7x-2 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=3,b_1=-8,c_1=4,a_2=4,b_2=-7,c_2=-2$
  We know that, condition for two quadratic equations having a common root is
  $(a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(4)(4)-(3)(-2)]^2 = [(3)(-7)-(4)(-8)] [(-8)(-2)-(-7)(4)]$
  or $[16+6]^2 = [-21+32] [16+28]$
  or $484 =484$
  Thus, $3x^2-8x+4=0$ and $4x^2-7x-2 =0$ has a common root.
Find the value of p so that each pair of the equations may have one root common
1. $4x^2+px-12=0$ and $4x^2+3px-4 =0$
  
  Solution 👉 Click Here
  
  Solution :2(a)
  Given quadratic equations are
  $4x^2+px-12=0$ and $4x^2+3px-4 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=4,b_1=p,c_1=-12,a_2=4,b_2=3p,c_2=-4$
  We know that, condition for two quadratic equations having a common root is
  $(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(-12)(4)-(-4)(4)]^2 = [(4)(3p)-(4)(p)] [(p)(-4)-(3p)(-12)]$
  or $[-48+16]^2 = [12p-4p] [-4p+36p]$
  or $[-32]^2 = [8p] [32p]$
  or $p^2=4$
  or $p=\pm 2$
  Thus,fo $p=\pm 2$ , given pair of the equations may have one root common
2. $2x^2+px-1=0$ and $3x^2-2x-5 =0$
  
  Solution 👉 Click Here
  
  Solution :2(b)
  Given quadratic equations are
  $2x^2+px-1=0$ and $3x^2-2x-5=0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=2,b_1=p,c_1=-1,a_2=3,b_2=-2,c_2=-5$
  We know that, condition for two quadratic equations having a common root is
  $(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(-1)(3)-(-5)(2)]^2 = [(2)(-2)-(3)(p)] [(p)(-5)-(-2)(-1)]$
  or $[-3+10]^2 = [-4-3p] [-5p-2]$
  or $[7]^2 = [4+3p] [2+5p]$
  or $49 = 15p^2+26p+8$
  or $15p^2+26p-41=0$
  or $15p^2+(41-15)p-41=0$
  or $15p^2+41p-15p-41=0$
  or $p(15p+41)-1(15p+41)=0$
  or $p=1, -\frac{41}{15}$
  Thus,fo $p=1, -\frac{41}{15}$ , given pair of the equations may have one root common
If the quadratic equations $x^2+px+q=0$ and $x^2+p'x+q'=0$ have common roots show that it must be either $\frac{pq'-p'q}{q-q'}$ or $\frac{q-q'}{p'-p}$

Solution 👉 Click Here

Solution :3
Given quadratic equations are
$x^2+px+q=0$ and $x^2+p'x+q'=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=1,b_1=p,c_1=q,a_2=1,b_2=p',c_2=q'$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(q)(1)-(q')(1)]^2 = [(1)(p')-(1)(p)] [(p)(q')-(p')(q)]$
or $[q-q']^2 = [p'-p] [pq'-p'q]$
or $\frac{pq'-p'q}{q-q'} . \frac{p'-p}{q-q'} =0$
or $\frac{pq'-p'q}{q-q'} =0$ or $\frac{p'-p}{q-q'} =0$
or $\frac{pq'-p'q}{q-q'} =0$ or $\frac{q-q'}{p'-p} =0$
Thus, $x^2+px+q=0$ and $x^2+p'x+q'=0$ have one root common if $\frac{pq'-p'q}{q-q'} =0$ or $\frac{q-q'}{p'-p} =0$
If the quadratic equations $x^2+px+q=0$ and $x^2+qx+p=0$ have common roots show that it must be either $p=q$ or $p+q+1=0$

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Solution :4
Given quadratic equations are
$x^2+px+q=0$ and $x^2+qx+p=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=1,b_1=p,c_1=q,a_2=1,b_2=q,c_2=p$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(q)(1)-(p)(1)]^2 = [(1)(q)-(1)(p)] [(p)(p)-(q)(q)]$
or $[q-p]^2 = [q-p] [p^2-q^2]$
or $(q-p) [(q-p)- (p^2-q^2)]=0$
or $q-p =0$ or $p+q-1 =0$
Thus, $x^2+px+q=0$ and $x^2+qx+p=0$ have one root common if $p=q$ or $p+q-1 =0$
If the quadratic equations $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have common roots show that it must be either $a=b=c$ or $a+b+c=0$

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Solution :5
Given quadratic equations are
$ax^2+bx+c=0$ and $bx^2+cx+a=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=a,b_1=b,c_1=c,a_2=b,b_2=c,c_2=a$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(c)(b)-(a)(a)]^2 = [(a)(c)-(b)(b)] [(b)(a)-(c)(c)]$
or $[bc-a^2]^2 = [ca-b^2] [ab-c^2]$
or $b^2c^2-2a^2bc+a^4=a^2bc-ac^3-ab^3+b^2c^2$
or $a^4=3a^2bc-ac^3-ab^3$
or $a^3=3abc-c^3-b^3$
or $a^3+b^3+c^3-3abc=0$
or $a^3+b^3+c^3-3abc=0$
or $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
or Either $(a+b+c)=0$ or $(a^2+b^2+c^2-ab-bc-ca)=0$
or Either $a+b+c=0$ or $2a^2+2b^2+2c^2-2ab-2bc-2ca=0$
or Either $a+b+c=0$ or $(a-b)^2+(b-c)^2+(c-a)^2=0$
or Either $a+b+c=0$ or $a-b=0, b-c=0,c-a$
or Either $a+b+c=0$ or $a=b,b=c,c=a$
or Either $a+b+c=0$ or $a=b=c$
Thus, $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have one root common if $a+b+c=0$ or $a=b=c$
Prove that if the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root, their other root will satisfy $x^2+ax+bc=0$

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Solution :6
Given quadratic equations are
$x^2+bx+ca=0$ and $x^2+cx+ab=0$
Let $x$ be the common solution of the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ then
$\frac{x^2}{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} }=\frac{x}{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}=\frac{1}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}}$
Comparing first two, we get
$x=\frac{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} } {\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix} }$
or $x=\frac{ab^2-ac^2}{ac-ab}$
or $x=-(b+c)$
Again, comparing last two, we get
$x= \frac{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}}$
or $x=\frac{ac-ab}{c-b}$
or $x=a$
Using both values of x, we get
$a=-(b+c)$
or $a+b+c=0$
Now, suppose $\alpha$ be the other root of $x^2+bx+ca=0$ , then
$x+\alpha=-b$
or $a+\alpha=-b$
or $\alpha=-a-b$
Next, suppose $\beta$ be the other root of $x^2+cx+ab=0$ , then
$x+\beta=-c$
or $a+\beta=-c$
or $\beta=-a-c$
Now, new quadratic equation containing other roots $\alpha, \beta$ is given by
$x^2-(\alpha+ \beta)x+ (\alpha \beta)=0$
or $x^2-(-a-b-a-c)x+ (a-b)(a-c)=0$
or $x^2-(-2a-b-c)x+ (a^2-ab-ac+bc)=0$
or $x^2-(-2a+a)x+ (0+bc)=0$
or $x^2+ax+ bc=0$