Mathematics 11
- Chapter 1 Sets
- Chapter 2 Relations And Functions
- Chapter 3 Trigonometric Functions
- Chapter 4 Complex numbers and quadratic equations
- Chapter 5 Linear Inequalities
- Chapter 6 Permutations And Combinations
- Chapter 7 Binomial Theorem
- Chapter 8 Sequences And Series
- Chapter 9 Straight Lines
- Chapter 10 Conic Sections
- Chapter 11 Introduction To Three Dimensional Geometry
- Chapter 12 Limits And Derivaties
- Chapter 13 Statistics
- Chapter 14 Probability
Chapter 8 Sequences And Series
SEQUENCES AND SERIES
Natural numbers are the product of human spirit. - DEDEKIND
8.1 Introduction
In mathematics, the word, “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on. For example, population of human beings or bacteria at different times form a sequence. The amount of money deposited in a bank, over a number of years form a sequence. Depreciated values of certain commodity occur in a sequence. Sequences have important applications in several spheres of human activities.
Sequences, following specific patterns are called progressions. In previous class, we have studied about arithmetic progression (A.P). In this Chapter, besides discussing more about A.P.; arithmetic mean, geometric mean, relationship between A.M. and G.M., special series in forms of sum to $n$ terms of consecutive natural numbers, sum to $n$ terms of squares of natural numbers and sum to $n$ terms of cubes of natural numbers will also be studied.
8.2 Sequences
Let us consider the following examples:
Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years.
Here, the total number of generations $=\frac{300}{30}=10$
The number of person’s ancestors for the first, second, third, …, tenth generations are $2,4,8,16,32, \ldots, 1024$. These numbers form what we call a sequence.
Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get $3,3.3,3.33,3.333, \ldots$ and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by $a_1, a_2, a_3, \ldots, a_n, \ldots$, etc., the subscripts denote the position of the term. The $n^{\text{th }}$ term is the number at the $n^{\text{th }}$ position of the sequence and is denoted by $a_n$. The $n^{\text{th }}$ term is also called the general term of the sequence.
Thus, the terms of the sequence of person’s ancestors mentioned above are:
$ a_1=2, a_2=4, a_3=8, \ldots, a _{10}=1024 $
Similarly, in the example of successive quotients
$ a_1=3, a_2=3.3, a_3=3.33, \ldots, a_6=3.33333 \text{, etc. } $
A sequence containing finite number of terms is called a finite sequence. For example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed number).
A sequence is called infinite, if it is not a finite sequence. For example, the sequence of successive quotients mentioned above is an infinite sequence, infinite in the sense that it never ends.
Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers $2,4,6, \ldots$
$ \begin{aligned} & \text{ Here } \quad a_1=2=2 \times 1 \quad a_2=4=2 \times 2 \\ & a_3=6=2 \times 3 \quad a_4=8=2 \times 4 \\ &\ldots & \ldots . & \ldots . & \ldots . & \ldots . & \ldots\\ & a _{23}=46=2 \times 23, a _{24}=48=2 \times 24 \text{, and so on. } \end{aligned} $
$ \begin{array}{lll} \text{Here } & a_1=2=2 \times 1 & a_2=4=2 \times 2 \\ & a_3=6=2 \times 3 & a_4=8=2 \times 4 \\ & \ldots \quad \dots \quad \dots & \ldots \quad \dots \quad \dots \\ & \ldots \quad \dots \quad \dots & \ldots \quad \dots \quad \dots \\ \end{array} \\ \quad \quad\quad a_{23}=46=2 \times 23, a_{24}=48=2 \times 24, \text{and so on} $
In fact, we see that the $n^{\text{th }}$ term of this sequence can be written as $a_n=2 n$, where $n$ is a natural number. Similarly, in the sequence of odd natural numbers $1,3,5, \ldots$, the $n^{\text{th }}$ term is given by the formula, $a_n=2 n-1$, where $n$ is a natural number.
In some cases, an arrangement of numbers such as $1,1,2,3,5,8, .$. has no visible pattern, but the sequence is generated by the recurrence relation given by
$ \begin{aligned} & a_1=a_2=1 \\ & a_3=a_1+a_2 \\ & a_n=a _{n-2}+a _{n-1}, n>2 \end{aligned} $
This sequence is called Fibonacci sequence.
In the sequence of primes $2,3,5,7, \ldots$, we find that there is no formula for the $n^{\text{th }}$ prime. Such sequence can only be described by verbal description.
In every sequence, we should not expect that its terms will necessarily be given by a specific formula. However, we expect a theoretical scheme or a rule for generating the terms $a_1, a_2, a_3, \ldots, a_n, \ldots$ in succession.
In view of the above, a sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for $a_n$.
8.3 Series
Let $a_1, a_2, a_3, \ldots, a_n$, be a given sequence. Then, the expression
$ a_1+a_2+a_3+\ldots+a_n+\ldots $
is called the series associated with the given sequence. The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter $\sum$ (sigma) as means of indicating the summation involved. Thus, the series $a_1+a_2+a_3+\ldots+a_n$ is abbreviated
as $\sum_{k=1}^{n} a_k$.
Remark When the series is used, it refers to the indicated sum not to the sum itself. For example, $1+3+5+7$ is a finite series with four terms. When we use the phrase “sum of a series,” we will mean the number that results from adding the terms, the sum of the series is 16 .
We now consider some examples.
Example 1 Write the first three terms in each of the following sequences defined by the following:
(i) $a_n=2 n+5$,
(ii) $a_n=\frac{n-3}{4}$.
Solution (i) Here $a_n=2 n+5$
Substituting $n=1,2,3$, we get
$ a_1=2(1)+5=7, a_2=9, a_3=11 $
Therefore, the required terms are 7, 9 and 11 .
(ii) Here $a_n=\frac{n-3}{4}$. Thus, $a_1=\frac{1-3}{4}=-\frac{1}{2}, a_2=-\frac{1}{4}, a_3=0$
Hence, the first three terms are $-\frac{1}{2},-\frac{1}{4}$ and 0 .
Example 2 What is the $20^{\text{th }}$ term of the sequence defined by
$ a_n=(n-1)(2-n)(3+n) ? $
Solution Putting $n=20$, we obtain
$ \begin{aligned} a _{20} & =(20-1)(2-20)(3+20) \\ & =19 \times(-18) \times(23)=-7866 . \end{aligned} $
Example 3 Let the sequence $a_n$ be defined as follows:
$ a_1=1, a_n=a _{n-1}+2 \text{ for } n \geq 2 \text{. } $
Find first five terms and write corresponding series.
Solution We have
$ \begin{aligned} & a_1=1, a_2=a_1+2=1+2=3, a_3=a_2+2=3+2=5, \\ & a_4=a_3+2=5+2=7, a_5=a_4+2=7+2=9 . \end{aligned} $
Hence, the first five terms of the sequence are $1,3,5,7$ and 9 . The corresponding series is $1+3+5+7+9+\ldots$
EXERCISE 8.1
Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text{th }}$ terms are:
1. $a_n=n(n+2)$
Answer : $a_n=n(n+2)$ Substituting $n=1,2,3,4$, and 5 , we obtain $a_1=1(1+2)=3$ $a_2=2(2+2)=8$ $a_3=3(3+2)=15$ $a_4=4(4+2)=24$ $a_5=5(5+2)=35$ Therefore, the required terms are $3,8,15,24$, and 35 .Show Answer
Answer : $a_n=\frac{n}{n+1}$ Substituting $n=1,2,3,4,5$, we obtain $a_1=\frac{1}{1+1}=\frac{1}{2}, a_2=\frac{2}{2+1}=\frac{2}{3}, a_3=\frac{3}{3+1}=\frac{3}{4}, a_4=\frac{4}{4+1}=\frac{4}{5}, a_5=\frac{5}{5+1}=\frac{5}{6}$ Therefore, the required terms are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}$, and $\frac{5}{6}$.Show Answer
Answer : $a_n=2^{n}$ Substituting $n=1,2,3,4,5$, we obtain $a_1=2^{1}=2$ $a_2=2^{2}=4$ $a_3=2^{3}=8$ $a_4=2^{4}=16$ $a_5=2^{5}=32$ Therefore, the required terms are 2, 4, 8, 16, and 32 .Show Answer
Answer : Substituting $n=1,2,3,4,5$, we obtain $a_1=\frac{2 \times 1-3}{6}=\frac{-1}{6}$ $a_2=\frac{2 \times 2-3}{6}=\frac{1}{6}$ $a_3=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$ $a_4=\frac{2 \times 4-3}{6}=\frac{5}{6}$ $a_5=\frac{2 \times 5-3}{6}=\frac{7}{6}$ Therefore, the required terms are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$, and $\frac{7}{6}$Show Answer
Answer : Substituting $n=1,2,3,4,5$, we obtain $a_1=(-1)^{1-1} 5^{1+1}=5^{2}=25$ $a_2=(-1)^{2-1} 5^{2+1}=-5^{3}=-125$ $a_3=(-1)^{3-1} 5^{3+1}=5^{4}=625$ $a_4=(-1)^{4-1} 5^{4+1}=-5^{5}=-3125$ $a^{5}=(-1)^{5-1} 5^{5+1}=5^{6}=15625$ Therefore, the required terms are 25, 125, 625, 3125, and 15625.Show Answer
Answer : Substituting $n=1,2,3,4,5$, we obtain $
\begin{aligned}
& a_1=1 \cdot \frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2} \\
& a_2=2 \cdot \frac{2^{2}+5}{4}=2 \cdot \frac{9}{4}=\frac{9}{2} \\
& a_3=3 \cdot \frac{3^{2}+5}{4}=3 \cdot \frac{14}{4}=\frac{21}{2} \\
& a_4=4 \cdot \frac{4^{2}+5}{4}=21 \\
& a_5=5 \cdot \frac{5^{2}+5}{4}=5 \cdot \frac{30}{4}=\frac{75}{2}
\end{aligned}
$ Therefore, the required terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$, and $\frac{75}{2}$. Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text{th }}$ terms are:Show Answer
Answer : Substituting $n=17$, we obtain $a _{17}=4(17)-3=68-3=65$ Substituting $n=24$, we obtain $a _{24}=4(24)-3=96-3=93$Show Answer
Answer : Substituting $n=7$, we obtain $a_7=\frac{7^{2}}{2 \times 7}=\frac{7}{2}$Show Answer
Answer : Substituting $n=9$, we obtain $a_9=(-1)^{9-1}(9)^{3}=(9)^{3}=729$Show Answer
Answer : Substituting $n=20$, we obtain $a _{20}=\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}$ Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:Show Answer
Answer : $a_1=3, a_n=3 a _{n-1}+2$ for all $n>1$ $\Rightarrow a_2=3 a_1+2=3(3)+2=11$ $a_3=3 a_2+2=3(11)+2=35$ $a_4=3 a_3+2=3(35)+2=107$ $a_5=3 a_4+2=3(107)+2=323$ Hence, the first five terms of the sequence are $3,11,35,107$, and 323 . The corresponding series is $3+11+35+107+323+\ldots$Show Answer
Answer : $a_1=-1, a_n=\frac{a _{n-1}}{n}, n \geq 2$ $\Rightarrow a_2=\frac{a_1}{2}=\frac{-1}{2}$ $a_3=\frac{a_2}{3}=\frac{-1}{6}$ $a_4=\frac{a_3}{4}=\frac{-1}{24}$ $a_5=\frac{a_4}{4}=\frac{-1}{120}$ Hence, the first five terms of the sequence are $
-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24} \text{, and } \frac{-1}{120}
$ The corresponding series is $
(-1)+(\frac{-1}{2})+(\frac{-1}{6})+(\frac{-1}{24})+(\frac{-1}{120})+\ldots
$Show Answer
Answer : $a_1=a_2=2, a_n=a _{n-1}-1, n>2$ $\Rightarrow a_3=a_2-1=2-1=1$ $a_4=a_3-1=1-1=0$ $a_5=a_4-1=0-1=-1$ Hence, the first five terms of the sequence are 2, 2, 1, 0 , and -1. The corresponding series is $(2+2+1+0- 1)+\ldots$Show Answer
$ 1=a_1=a_2 \text{ and } a_n=a _{n-1}+a _{n-2}, n>2 \text{. } $
Find $\frac{a _{n+1}}{a_n}$, for $n=1,2,3,4,5$
Show Answer
Answer :
$1=a_1=a_2$
$a_n=a _{n-1}+a _{n-2}, n>2$
$\therefore a_3=a_2+a_1=1+l=2$
$a_4=a_3+a_2=2+1=3$
$a_5=a_4+a_3=3+2=5$
$a_6=a_5+a_4=5+3=8$
$\therefore$ For $n=1, \frac{a_n+1}{a_n}=\frac{a_2}{a_1}=\frac{1}{1}=1$
For $n=2, \frac{a_n+1}{a_n}=\frac{a_3}{a_2}=\frac{2}{1}=2$
For $n=3, \frac{a_n+1}{a_n}=\frac{a_4}{a_3}=\frac{3}{2}$
For $n=4, \frac{a_n+1}{a_n}=\frac{a_5}{a_4}=\frac{5}{3}$
For $n=5, \frac{a_n+1}{a_n}=\frac{a_6}{a_5}=\frac{8}{5}$
8.4 Geometric Progression (G. P.)
Let us consider the following sequences:
(i) $2,4,8,16, \ldots$,
(ii) $\frac{1}{9}, \frac{-1}{27}, \frac{1}{81}, \frac{-1}{243}$
(iii) $.01, .0001, .000001, \ldots$
In each of these sequences, how their terms progress? We note that each term, except the first progresses in a definite order.
In (i), we have $a_1=2, \frac{a_2}{a_1}=2, \frac{a_3}{a_2}=2, \frac{a_4}{a_3}=2$ and so on.
In (ii), we observe, $a_1=\frac{1}{9}, \frac{a_2}{a_1}=\frac{1}{3}, \frac{a_3}{a_2}=\frac{1}{3}, \frac{a_4}{a_3}=\frac{1}{3}$ and so on.
Similarly, state how do the terms in (iii) progress? It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it. In (i), this constant ratio is 2; in (ii), it is $-\frac{1}{3}$ and in (iii), the constant ratio is 0.01 . Such sequences are called geometric sequence or geometric progression abbreviated as G.P.
A sequence $a_1, a_2, a_3, \ldots, a_n, \ldots$ is called geometric progression, if each term is non-zero and $\frac{a_k+1}{a_k}=r$ (constant), for $k \geq 1$.
By letting $a_1=a$, we obtain a geometric progression, $a, a r, a r^{2}, a r^{3}, \ldots$, where $a$ is called the first term and $r$ is called the common ratio of the G.P. Common ratio in geometric progression (i), (ii) and (iii) above are $2,-\frac{1}{3}$ and 0.01 , respectively.
As in case of arithmetic progression, the problem of finding the $n^{\text{th }}$ term or sum of $n$ terms of a geometric progression containing a large number of terms would be difficult without the use of the formulae which we shall develop in the next Section. We shall use the following notations with these formulae:
$ \begin{aligned} & a=\text{ the first term, } r=\text{ the common ratio, } l=\text{ the last term, } \\ & n=\text{ the numbers of terms, } \\ & S_n=\text{ the sum of first } n \text{ terms. } \end{aligned} $
8.4.1 General term of $a$ G.P.
Let us consider a G.P. with first non-zero term ’ $a$ ’ and common ratio ’ $r$ ‘. Write a few terms of it. The second term is obtained by multiplying $a$ by $r$, thus $a_2=a r$. Similarly, third term is obtained by multiplying $a_2$ by $r$. Thus, $a_3=a_2 r=a r^{2}$, and so on.
We write below these and few more terms.
$1^{\text{st }}$ term $=a_1=a=a r^{1-1}, 2^{\text{nd }}$ term $=a_2=a r=a r^{2-1}, 3^{\text{rd }}$ term $=a_3=a r^{2}=a r^{3-1}$
$4^{\text{th }}$ term $=a_4=a r^{3}=a r^{4-1}, 5^{\text{th }}$ term $=a_5=a r^{4}=a r^{5-1}$
Do you see a pattern? What will be $16^{\text{th }}$ term?
$ a _{16}=a r^{16-1}=a r^{15} $
Therefore, the pattern suggests that the $n^{\text{th }}$ term of a G.P. is given by $a_n=a r^{n-1}$.
Thus, $a$, G.P. can be written as $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1} ; a, a r, a r^{2}, \ldots, a r^{n-1} \ldots ;$ according as G.P. is finite or infinite, respectively.
The series $a+a r+a r^{2}+\ldots+a r^{n-1}$ or $a+a r+a r^{2}+\ldots+a r^{n-1}+\ldots$ are called finite or infinite geometric series, respectively.
8.4.2. Sum to $n$ terms of $a$ G.P.
Let the first term of a G.P. be $a$ and the common ratio be $r$. Let us denote by $S_n$ the sum to first $n$ terms of G.P. Then
$ S_n=a+a^{n}+a r^{2}+\ldots+a r^{n-1} \quad \quad \quad \quad \quad \ldots (1) $
Case 1 If $r=1$, we have $S_n=a+a+a+\ldots+a(n$ terms $)=n a$
Case 2 If $r \neq 1$, multiplying (1) by $r$, we have
$ r S_n=a r+a r^{2}+a r^{3}+\ldots+a r^{n} \quad \quad \quad \quad \quad \ldots (2) $
Subtracting (2) from (1), we get $(1-r) S_n=a-a r^{n}=a(1-r^{n})$
This gives
$ \text{ or } \quad S_n=\frac{a(r^{n}-1)}{r-1} $
Example 4 Find the $10^{\text{th }}$ and $n^{\text{th }}$ terms of the G.P. $5,25,125, \ldots$.
Solution Here $a=5$ and $r=5$. Thus, $a _{10}=5(5)^{10-1}=5(5)^{9}=5^{10}$ and $a_n=a r^{n-1}=5(5)^{n-1}=5^{n}$.
Example 5 Which term of the G.P., 2,8,32,… up to $n$ terms is 131072 ?
Solution Let 131072 be the $n^{\text{th }}$ term of the given G.P. Here $a=2$ and $r=4$.
Therefore $\quad 131072=a_n=2(4)^{n-1}$ or $65536=4^{n-1}$
This gives $\quad 4^{8}=4^{n-1}$.
So that $n-1=8$, i.e., $n=9$. Hence, 131072 is the $9^{\text{th }}$ term of the G.P.
Example 6 In a G.P., the $3^{\text{rd }}$ term is 24 and the $6^{\text{th }}$ term is 192.Find the $10^{\text{th }}$ term.
Solution Here, $a_3=a r^{2}=24 \quad \quad \quad \quad \quad \ldots (1)$
and $ \quad \quad a_6=a r^{5}=192 \quad \quad \quad \quad \quad \ldots (2) $
Dividing (2) by (1), we get $r=2$. Substituting $r=2$ in (1), we get $a=6$.
Hence $a _{10}=6(2)^{9}=3072$.
Example 7 Find the sum of first $n$ terms and the sum of first 5 terms of the geometric series $1+\frac{2}{3}+\frac{4}{9}+\ldots$
Solution Here $a=1$ and $r=\frac{2}{3}$. Therefore
$ S_n=\frac{a(1-r^{n})}{1-r}=\frac{[1-(\frac{2}{3})^{n}]}{1-\frac{2}{3}}=3[1-(\frac{2}{3})^{n}] $
In particular, $\quad S_5=3[1-(\frac{2}{3})^{5}]=3 \times \frac{211}{243}=\frac{211}{81}$.
Example 8 How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the $sum \frac{3069}{512} ?$
Solution Let $n$ be the number of terms needed. Given that $a=3, r=\frac{1}{2}$ and $S_n=\frac{3069}{512}$
Since $ \quad \quad \quad S_n=\frac{a(1-r^{n})}{1-r} $
Therefore $ \quad \quad \quad \frac{3069}{512}=\frac{3(1-\frac{1}{2^{n}})}{1-\frac{1}{2}}=6(1-\frac{1}{2^{n}}) $
or $ \quad \quad \quad \frac{3069}{3072}=1-\frac{1}{2^{n}} $
or $\quad \quad \quad \frac{1}{2^{n}} =1-\frac{3069}{3072}=\frac{3}{3072}=\frac{1}{1024}$
or $\quad \quad \quad2^{n} =1024=2^{10}, \text{ which gives } n=10$
Example 9 The sum of first three terms of a G.P. is $\frac{13}{12}$ and their product is -1 .
Find the common ratio and the terms.
Solution Let $\frac{a}{r}, a$, ar be the first three terms of the G.P. Then
$ \frac{a}{r}+a r+a=\frac{13}{12} \quad \quad \quad \quad \quad \ldots (1) $
and $\quad(\frac{a}{r})(a)(a r)=-1 \quad \quad \quad \quad \quad \ldots (2) $
From (2), we get $a^{3}=-1$, i.e., $a=-1$ (considering only real roots)
Substituting $a=-1$ in (1), we have
$ -\frac{1}{r}-1-r=\frac{13}{12} \text{ or } 12 r^{2}+25 r+12=0 \text{. } $
This is a quadratic in $r$, solving, we get $r=-\frac{3}{4}$ or $-\frac{4}{3}$.
Thus, the three terms of G.P. are : $\frac{4}{3},-1, \frac{3}{4}$ for $r=\frac{-3}{4}$ and $\frac{3}{4},-1, \frac{4}{3}$ for $r=\frac{-4}{3}$,
Example10 Find the sum of the sequence 7, 77, 777, 7777, … to $n$ terms.
Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as
$ S_n=7+77+777+7777+\ldots \text{ to } n \text{ terms } $
$ \begin{aligned} & =\frac{7}{9}[9+99+999+9999+\ldots \text{ to } n \text{ term }] \\ & =\frac{7}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+(10^{4}-1)+\ldots n \text{ terms }] \\ & =\frac{7}{9}[(10+10^{2}+10^{3}+\ldots n \text{ terms })-(1+1+1+\ldots n \text{ terms })] \\ & =\frac{7}{9} \left[ \frac{10(10^{n}-1)}{10-1}-n\right]=\frac{7}{9}\left[\frac{10(10^{n}-1)}{9}-n \right] . \end{aligned} $
Example 11 A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.
Solution Here $a=2, r=2$ and $n=10$
Using the sum formula $\quad S_n=\frac{a(r^{n}-1)}{r-1}$
We have $ \quad\quad\quad\quad S_{10}=2(2^{10}-1)=2046 $
Hence, the number of ancestors preceding the person is 2046.
8.4.3 Geometric Mean (G.M.)
The geometric mean of two positive numbers $a$ and $b$ is the number $\sqrt{a b}$. Therefore, the geometric mean of 2 and 8 is 4 . We observe that the three numbers $2,4,8$ are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.
Given any two positive numbers $a$ and $b$, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.
Let $G_1, G_2, \ldots, G_n$ be $n$ numbers between positive numbers $a$ and $b$ such that $a, G_1, G_2, G_3, \ldots, G_n, b$ is a G.P. Thus, $b$ being the $(n+2)^{\text{th }}$ term, we have
$ b=a r^{n+1}, \quad \text{ or } \quad r=(\frac{b}{a})^{\frac{1}{n+1}} \text{. } $
Hence $G_1=a r=a(\frac{b}{a})^{\frac{1}{n+1}}, G_2=a r^{2}=a(\frac{b}{a})^{\frac{2}{n+1}}, G_3=a r^{3}=a(\frac{b}{a})^{\frac{3}{n+1}}$,
$ G_n=a r^{n}=a(\frac{b}{a})^{\frac{n}{n+1}} $
Example12 Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Solution Let $G_1, G_2, G_3$ be three numbers between 1 and 256 such that $1, G_1, G_2, G_3, 256$ is a G.P.
Therefore $\quad 256=r^{4}$ giving $r= \pm 4$ (Taking real roots only)
For $r=4$, we have $G_1=a r=4, G_2=a r^{2}=16, G_3=a r^{3}=64$
Similarly, for $r=-4$, numbers are $-4,16$ and -64 .
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are in G.P.
8.5 Relationship Between A.M. and G.M.
Let $A$ and $G$ be A.M. and G.M. of two given positive real numbers $a$ and $b$, respectively. Then
$ A=\frac{a+b}{2} \text{ and } G=\sqrt{a b} $
Thus, we have
$ \begin{aligned} A-G & =\frac{a+b}{2}-\sqrt{a b}=\frac{a+b-2 \sqrt{a b}}{2} \\ & =\frac{(\sqrt{a}-\sqrt{b})^{2}}{2} \geq 0 \quad \quad \quad \quad \quad \quad \quad \ldots (1) \end{aligned} $
From (1), we obtain the relationship $A \geq G$.
Example 13 If A.M. and G.M. of two positive numbers $a$ and $b$ are 10 and 8, respectively, find the numbers.
Solution Given that A.M. $=\frac{a+b}{2}=10 \quad \quad \quad \quad \quad \quad \quad \ldots (1)$
and $ \text{ G.M. }=\sqrt{a b}=8 \quad \quad \quad \quad \quad \quad \quad \ldots (2) $
From (1) and (2), we get
$ \begin{aligned} & a+b=20 \quad \quad \quad \quad \quad \quad \quad \ldots (3)\\ & a b=64 \quad \quad \quad \quad \quad \quad \quad \ldots (4) \end{aligned} $
Putting the value of $a$ and $b$ from (3), (4) in the identity $(a-b)^{2}=(a+b)^{2}-4 a b$, we get
$(a-b)^{2}=400-256=144$
or $\quad \quad \quad a-b= \pm 12 \quad \quad \quad \quad \quad \quad \quad \ldots (5)$
Solving (3) and (5), we obtain
$ a=4, b=16 \text{ or } a=16, b=4 $
Thus, the numbers $a$ and $b$ are 4,16 or 16,4 respectively.
EXERCISE 8.2
1. Find the $20^{\text{th }}$ and $n^{\text{th }}$ terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$
Answer : The given G.P. is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$ Here, $a=$ First term $=\frac{5}{2}$ $$
\begin{gathered}
\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2} \\
r=\text{ Common ratio }= \\
a _{20}=a r^{20-1}=\frac{5}{2}(\frac{1}{2})^{19}=\frac{5}{(2)(2)^{19}}=\frac{5}{(2)^{20}} \\
a_n=a r^{n-1}=\frac{5}{2}(\frac{1}{2})^{n-1}=\frac{5}{(2)(2)^{n-1}}=\frac{5}{(2)^{n}}
\end{gathered}
$$Show Answer
Answer : Common ratio, $r=2$ Let $a$ be the first term of the G.P. $
\begin{aligned}
& \therefore a_8=a r^{8 e^{-1}}=a r^{7} \\
& \Rightarrow a r^{7}=192 \\
& a(2)^{7}=192 \\
& a(2)^{7}=(2)^{6}(3) \\
& \Rightarrow a=\frac{(2)^{6} \times 3}{(2)^{7}}=\frac{3}{2} \\
& \therefore a _{12}=a r^{12-1}=(\frac{3}{2})(2)^{11}=(3)(2)^{10}=3072
\end{aligned}
$Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. According to the given condition, $a_5=a r^{5 - 1}=a r^{4}=p$. $a_8=a r^{8 - 1}=a r^{7}=q \ldots(2)$ $a_{11}=a r^{11 -1}=a r^{10}=s \ldots$ Dividing equation (2) by (1), we obtain $\frac{a r^{7}}{a r^{4}}=\frac{q}{p}$ $r^{3}=\frac{q}{p}$ Dividing equation (3) by (2), we obtain $\frac{a r^{10}}{a r^{7}}=\frac{s}{q}$ $\Rightarrow r^{3}=\frac{s}{q}$ Equating the values of $r^{3}$ obtained in (4) and (5), we obtain $\frac{q}{p}=\frac{s}{q}$ $\Rightarrow q^{2}=p s$ Thus, the given result is proved.Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. $\therefore a=-3$ It is known that, $a_n=a r^{n-1}$ $\therefore a_4=a r^{3}=(-3) r^{3}$ $a_2=a r^{1}=(-3) r$ According to the given condition, $(-3) r^{3}=[(-3) r]^{2}$ $\Rightarrow-3 r^{3}=9 r^{2}$ $\Rightarrow r=-3$ $a_7=a r^{7.1}=a r^{6}=(-3)(-3)^{6}=-(3)^{7}=-2187$ Thus, the seventh term of the G.P. is -2187 .Show Answer
(a) $2,2 \sqrt{2}, 4, \ldots$ is 128 ?
(b) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is 729 ?
(c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$ ?
Answer : (a) The given sequence is $2,2 \sqrt{2}, 4, \ldots$ Here, $a=2$ and $r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$ Let the $n^{\text{th }}$ term of the given sequence be 128 . $a_n=a r^{n-1}$ $\Rightarrow(2)(\sqrt{2})^{n-1}=128$ $\Rightarrow(2)(2)^{\frac{n-1}{2}}=(2)^{7}$ $\Rightarrow(2)^{\frac{n-1}{2}+1}=(2)^{7}$ $\therefore \frac{n-1}{2}+1=7$ $\Rightarrow \frac{n-1}{2}=6$ $\Rightarrow n-1=12$ $\Rightarrow n=13$ Thus, the $13^{\text{th }}$ term of the given sequence is 128 . (b) The given sequence is $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ Here, $
a=\sqrt{3} \text{ and } r=\frac{3}{\sqrt{3}}=\sqrt{3}
$ Let the $n^{\text{th }}$ term of the given sequence be 729 .
$a_n=a r^{n-1}$ $\therefore a r^{n-1}=729$ $\Rightarrow(\sqrt{3})(\sqrt{3})^{n-1}=729$ $\Rightarrow(3)^{\frac{1}{2}}(3)^{\frac{n-1}{2}}=(3)^{6}$ $\Rightarrow(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^{6}$ $\therefore \frac{1}{2}+\frac{n-1}{2}=6$ $\Rightarrow \frac{1+n-1}{2}=6$ $\Rightarrow n=12$ Thus, the $12^{\text{th }}$ term of the given sequence is 729 . (c) The given sequence is $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ Here, $\quad a=\frac{1}{3}$ and $r=\frac{1}{9} \div \frac{1}{3}=\frac{1}{3}$ Let the $n^{\text{th }}$ term of the given sequence be $\frac{1}{19683}$. $a_n=a r^{n-1}$ $\therefore a r^{n-1}=\frac{1}{19683}$ $\Rightarrow(\frac{1}{3})(\frac{1}{3})^{n-1}=\frac{1}{19683}$ $\Rightarrow(\frac{1}{3})^{n}=(\frac{1}{3})^{9}$ $\Rightarrow n=9$ Thus, the $9^{\text{th }}$ term of the given sequence is $\frac{1}{19683}$.Show Answer
Answer : The given numbers are $\frac{-2}{7}, x, \frac{-7}{2}$. $\frac{x}{-2}=\frac{-7 x}{2}$ Common ratio $=7$ Also, common ratio $=\frac{\frac{-7}{2}}{x}=\frac{-7}{2 x}$ $\therefore \frac{-7 x}{2}=\frac{-7}{2 x}$ $\Rightarrow x^{2}=\frac{-2 \times 7}{-2 \times 7}=1$ $\Rightarrow x=\sqrt{1}$ $\Rightarrow x= \pm 1$ Thus, for $x= \pm 1$, the given numbers will be in G.P. Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10 :Show Answer
Answer : The given G.P. is $0.15,0.015,0.00015, \ldots$ Here, $a=0.15$ and $
r=\frac{0.015}{0.15}=0.1
$ $S_n=\frac{a(1-r^{n})}{1-r}$ $\therefore S _{20}=\frac{0.15[1-(0.1)^{20}]}{1-0.1}$ $
=\frac{0.15}{0.9}[1-(0.1)^{20}]
$ $
=\frac{15}{90}[1-(0.1)^{20}]
$ $=\frac{1}{6}[1-(0.1)^{20}]$Show Answer
Answer : The given G.P. is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$ Here, $a=\sqrt{7}$ $r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$ $S _{12}=\frac{a(1-r^{n})}{1-r}$ $\therefore S_n=\frac{\sqrt{7}[1-(\sqrt{3})^{n}]}{1-\sqrt{3}}$ $=\frac{\sqrt{7}[1-(\sqrt{3})^{n}]}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}$ (By rationalizing) $=\frac{\sqrt{7}(1+\sqrt{3})[1-(\sqrt{3})^{n}]}{1-3}$ $=\frac{-\sqrt{7}(1+\sqrt{3})}{2}[1-(3)^{\frac{n}{2}}]$ $=\frac{\sqrt{7}(1+\sqrt{3})}{2}[(3)^{\frac{n}{2}}-1]$Show Answer
Answer : The given G.P. is $1,-a, a^{2},-a^{3}$, Here, first term $=a_1=1$ Common ratio $=r=- a$ $
\begin{aligned}
& S_n=\frac{a_1(1-r^{n})}{1-r} \\
& \therefore S_n=\frac{1[1-(-a)^{n}]}{1-(-a)}=\frac{[1-(-a)^{n}]}{1+a}
\end{aligned}
$Show Answer
Answer : The given G.P. is $x^{3}, x^{5}, x^{7}, \ldots$ Here, $a=x^{3}$ and $r=x^{2}$ $S_n=\frac{a(1-r^{n})}{1-r}=\frac{x^{3}[1-(x^{2})^{n}]}{1-x^{2}}=\frac{x^{3}(1-x^{2 n})}{1-x^{2}}$Show Answer
Answer : $\sum _{k=1}^{11}(2+3^{k})=\sum _{k=1}^{11}(2)+\sum _{k=1}^{11} 3^{k}=2(11)+\sum _{k=1}^{11} 3^{k}=22+\sum _{k=1}^{11} 3^{k}$ $\sum _{k=1}^{11} 3^{k}=3^{1}+3^{2}+3^{3}+\ldots+3^{11}$ The terms of this sequence $3,3^{2}, 3^{3}, \ldots$ forms a G.P.
$S_n=\frac{a(r^{n}-1)}{r-1}$ $\Rightarrow S _{11}=\frac{3[(3)^{11}-1]}{3-1}$ $\Rightarrow S _{11}=\frac{3}{2}(3^{\prime \prime}-1)$ $\therefore \sum _{k=1}^{11} 3^{k}=\frac{3}{2}(3^{11}-1)$ Substituting this value in equation (1), we obtain $\sum _{k=1}^{11}(2+3^{k})=22+\frac{3}{2}(3^{11}-1)$Show Answer
Answer : Let $\frac{a}{r}, a, a r$ be the first three terms of the G.P. $\frac{a}{r}+a+a r=\frac{39}{10}$ $(\frac{a}{r})(a)(a r)=1$ From (2), we obtain $a^{3}=1$ $\Rightarrow a=1$ (Considering real roots only) Substituting $a=1$ in equation (1), we obtain
$\frac{1}{r}+1+r=\frac{39}{10}$ $\Rightarrow 1+r+r^{2}=\frac{39}{10} r$ $\Rightarrow 10+10 r+10 r^{2}-39 r=0$ $\Rightarrow 10 r^{2}-29 r+10=0$ $\Rightarrow 10 r^{2}-25 r-4 r+10=0$ $\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$ $\Rightarrow(5 r-2)(2 r-5)=0$ $\Rightarrow r=\frac{2}{5}$ or $\frac{5}{2}$ Thus, the three terms of G.P. are $\frac{5}{2}, 1$, and $\frac{2}{5}$.Show Answer
Answer : The given G.P. is $3,3^{2}, 3^{3}, \ldots$ Let $n$ terms of this G.P. be required to obtain the sum as 120 . $S_n=\frac{a(r^{n}-1)}{r-1}$ Here, $a=3$ and $r=3$ $\therefore S_n=120=\frac{3(3^{n}-1)}{3-1}$ $\Rightarrow 120=\frac{3(3^{n}-1)}{2}$ $\Rightarrow \frac{120 \times 2}{3}=3^{n}-1$ $\Rightarrow 3^{n}-1=80$ $\Rightarrow 3^{n}=81$ $\Rightarrow 3^{n}=3^{4}$ $\therefore n=4$ Thus, four terms of the given G.P. are required to obtain the sum as 120 .Show Answer
Answer : Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots$ According to the given condition, $a+a r+a r^{2}=16$ and $a r^{3}+a r^{4}+a r^{5}=128$ $\Rightarrow a(1+r+r^{2})=16$ $ar^{3}(1+r+r^{2})=128$ Dividing equation (2) by (1), we obtain $\frac{a r^{3}(1+r+r^{2})}{a(1+r+r^{2})}=\frac{128}{16}$ $\Rightarrow r^{3}=8$ $\therefore r=2$ Substituting $r=2$ in (1), we obtain $a(1+2+4)=16$ $\Rightarrow a(7)=16$ $\Rightarrow a=\frac{16}{7}$ $S_n=\frac{a(r^{n}-1)}{r-1}$ $\Rightarrow S_n=\frac{16}{7} \frac{(2^{n}-1)}{2-1}=\frac{16}{7}(2^{n}-1)$Show Answer
Answer : $a=729$ $a_7=64$ Let $r$ be the common ratio of the G.P. It is known that, $a_n=a r^{\text{nâe-1 }}$ $a_7=a^{r e^{-1}}=(729) r^{6}$ $\Rightarrow 64=729 r^{6}$ $\Rightarrow r^{6}=\frac{64}{729}$ $\Rightarrow r^{6}=(\frac{2}{3})^{6}$ $\Rightarrow r=\frac{2}{3}$ Also, it is known that, $S_n=\frac{a(1-r^{n})}{1-r}$ $
\begin{aligned}
\therefore S_7 & =\frac{729[1-(\frac{2}{3})^{7}]}{1-\frac{2}{3}} \\
& =3 \times 729[1-(\frac{2}{3})^{7}] \\
& =(3)^{7}[\frac{(3)^{7}-(2)^{7}}{(3)^{7}}] \\
& =(3)^{7}-(2)^{7} \\
& =2187-128 \\
& =2059
\end{aligned}
$Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. According to the given conditions, $S_2=-4=\frac{a(1-r^{2})}{1-r}$ $a_5=4 \times a_3$ $a r^{4}=4 a r^{2}$
$\Rightarrow r^{2}=4$ $\therefore r= \pm 2$ From (1), we obtain $-4=\frac{a[1-(2)^{2}]}{1-2}$ for $r=2$ $\Rightarrow-4=\frac{a(1-4)}{-1}$ $\Rightarrow-4=a(3)$ $\Rightarrow a=\frac{-4}{3}$ Also, $-4=\frac{a[1-(-2)^{2}]}{1-(-2)}$ for $r=-2$ $\Rightarrow-4=\frac{a(1-4)}{1+2}$ $\Rightarrow-4=\frac{a(-3)}{3}$ $\Rightarrow a=4$ Thus, the required G.P. is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots$ or 4, 8, 16, 32, …Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. According to the given condition, $a_4=a r^{3}=x$ $a _{10}=a r^{9}=y$.. $a _{16}=a r^{15}=z$ Dividing (2) by (1), we obtain $\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}$ Dividing (3) by (2), we obtain $\frac{z}{y}=\frac{a r^{15}}{a r^{9}} \Rightarrow \frac{z}{y}=r^{6}$ $\therefore \frac{y}{x}=\frac{z}{y}$ Thus, $x, y, z$ are in G. P.Show Answer
Answer : The given sequence is $8,88,888,8888 \ldots$ This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as $S_n=8+88+888+8888+$ to $n$ terms $=\frac{8}{9}[9+99+999+9999+$ ..to $n$ terms] $=\frac{8}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+(10^{4}-1)+\ldots \ldots ..$. to $n$ terms $]$ $=\frac{8}{9}[(10+10^{2}+\ldots . . n..$ terms $)-(1+1+1+\ldots . n$ terms $.)]$ $=\frac{8}{9}[\frac{10(10^{n}-1)}{10-1}-n]$ $=\frac{8}{9}[\frac{10(10^{n}-1)}{9}-n]$ $=\frac{80}{81}(10^{n}-1)-\frac{8}{9} n$Show Answer
16,32 and $128,32,8,2, \frac{1}{2}$.
Answer : Required sum $=2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \frac{1}{2}$ $=64[4+2+1+\frac{1}{2}+\frac{1}{2^{2}}]$ Here, $4,2,1, \frac{1}{2}, \frac{1}{2^{2}}$ is a G.P. First term, $a=4$ Common ratio, $r=\frac{1}{2}$ It is known that, $S_n=\frac{a(1-r^{n})}{1-r}$ $\therefore S_5=\frac{4[1-(\frac{1}{2})^{5}]}{1-\frac{1}{2}}=\frac{4[1-\frac{1}{32}]}{\frac{1}{2}}=8(\frac{32-1}{32})=\frac{31}{4}$ $\therefore$ Required sum $=$ $ 64(\frac{31}{4})=(16)(31)=496
$Show Answer
Answer : It has to be proved that the sequence, $a A, a r A R, a r^{2} A R^{2}, \ldots a r^{n -1} A R^{n - 1}$, forms a G.P. $
\begin{aligned}
& \frac{\text{ Second term }}{\text{ First term }}=\frac{a r A R}{a A}=r R \\
& \frac{\text{ Third term }}{\text{ Second term }}=\frac{a r^{2} A R^{2}}{a r A R}=r R
\end{aligned}
$ Thus, the above sequence forms a G.P. and the common ratio is $r R$.Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. $a_1=a, a_2=a r, a_3=a r^{2}, a_4=a r^{3}$ By the given condition, $a_3=a_1+9$ $\Rightarrow a r^{2}=a+9$ $a_2=a_4+18$ $\Rightarrow a r=a r^{3}+18$. From (1) and (2), we obtain $a(r^{2}- 1)=9$ $ar(1-{2})=18$ Dividing (4) by (3), we obtain $\frac{a r(1-r^{2})}{a(r^{2}-1)}=\frac{18}{9}$ $\Rightarrow-r=2$ $\Rightarrow r=-2$ Substituting the value of $r$ in (1), we obtain $4 a=a+9$ $\Rightarrow 3 a=9$ $\therefore a=3$ Thus, the first four numbers of the G.P. are 3, 3(2), 3(2)2, and 3 (2) i.e., 3 ,6, 12 , and 24.Show Answer
$ a^{q-r} b^{r-p} c^{P-q}=1 $
Answer : Let $A$ be the first term and $R$ be the common ratio of the G.P. According to the given information, $A R^{p-1}=a$ $A R^{q-1}=b$ $A R^{r-1}=c$ $a^{q-r} b^{r-p} C^{p-q}$
$=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}$ $=A q^{-r+r-p+p-q} \times R^{(p r-p r-q+r+(r q-r+p-p q)+(p r-p-q r+q)}$ $=A^{0} \times R^{0}$ $=1$ Thus, the given result is proved.Show Answer
Answer : The first term of the G.P is $a$ and the last term is $b$. Therefore, the G.P. is $a, a r, a r^{2}, a r^{3}, \ldots a r^{n \in E 1}$, where $r$ is the common ratio. $b=a^{n n^{n} e^{-1}}$. $P=$ Product of $n$ terms $=(a)(a r)(a r^{2}) \ldots(a r^{n E^{n}-1})$ $=(a \times a \times \ldots a)(r \times r^{2} \times \ldots r^{r^{n} e^{-1}})$ $=a^{n} r^{1+2+\ldots(n a E-1)}$ Here, $1,2, \ldots(n \hat{a}.$ “" $.^{\prime})$ is an A.P. $\therefore 1+2+\ldots \ldots \ldots+(n- 1)=\frac{n-1}{2}[2+(n-1-1) \times 1]=\frac{n-1}{2}[2+n-2]=\frac{n(n-1)}{2}$ $P=a^{n} r^{\frac{n(n-1)}{2}}$ $\therefore P^{2}=a^{2 n} r^{n(n-1)}$ $=[a^{2} r^{(n-1)}]^{n}$ $=[a \times ar^{n-1}]^{n}$ $=(ab)^{n}$ $[U \sin g(1)]$ Thus, the given result is proved.Show Answer
Answer : Let $a$ be the first term and $r$ be the common ratio of the G.P. Sum of first $n$ terms $=\frac{a(1-r^{n})}{(1-r)}$ Since there are $n$ terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term, Sum of terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term $
=\frac{a _{n+1}(1-r^{n})}{(1-r)}
$ $a^{n+1}=a r^{n+1-1}=a r^{n}$ Thus, required ratio $=\frac{a(1-r^{n})}{(1-r)} \times \frac{(1-r)}{a^{n}(1-r^{n})}=\frac{1}{r^{n}}$ Thus, the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term is $\frac{1}{r^{n}}$.Show Answer
Answer : $a, b, c, d$ are in G.P. Therefore, $b c=a d$. $b^{2}=a c$. $c^{2}=b d$. It has to be proved that, $(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(a b+b c \text{ - } c d)^{2}$ R.H.S. $=(a b+b c+c d)^{2}$ $=(a b+a d+c d)^{2}[$ Using (1)] $=[a b+d(a+c)]^{2}$ $
\begin{aligned}
& =a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2} \\
& =a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}(a^{2}+2 a c+c^{2}) \\
& =a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}[\text{ Using (1) and (2)] } \\
& =a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2} \\
& =a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} \times b^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} b^{2}+c^{2} \times c^{2}+c^{2} d^{2}
\end{aligned}
$ [Using (2) and (3) and rearranging terms] $=a^{2}(b^{2}+c^{2}+d^{2})+b^{2}(b^{2}+c^{2}+d^{2})+c^{2}(b^{2}+c^{2}+d^{2})$ $=(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})$ $=$ L.H.S. $\therefore$ L.H.S. $=$ R.H.S. $\therefore(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(a b+b c+c d)^{2}$Show Answer
Answer : Let $G_1$ and $G_2$ be two numbers between 3 and 81 such that the series, $3, G_1, G_2$, 81, forms a G.P. Let $a$ be the first term and $r$ be the common ratio of the G.P. $\therefore 81=(3)(r)^{3}$ $\Rightarrow r^{3}=27$ $\therefore r=3$ (Taking real roots only) For $r=3$, $G_1=a r=(3)(3)=9$ $G_2=a r^{2}=(3)(3)^{2}=27$ Thus, the required two numbers are 9 and 27.Show Answer
Answer : G. M. of $a$ and $b$ is $\sqrt{a b}$. By the given condition, $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b}$ Squaring both sides, we obtain $
\frac{(a^{n+1}+b^{n+1})^{2}}{(a^{n}+b^{n})^{2}}=a b
$ $\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=(a b)(a^{2 n}+2 a^{n} b^{n}+b^{2 n})$ $\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=a^{2 n+1} b+2 a^{n+1} b^{n+1}+a b^{2 n+1}$ $\Rightarrow a^{2 n+2}+b^{2 n+2}=a^{2 n+1} b+a b^{2 n+1}$ $\Rightarrow a^{2 n+2}-a^{2 n+1} b=a b^{2 n+1}-b^{2 n+2}$ $\Rightarrow a^{2 n+1}(a-b)=b^{2 n+1}(a-b)$ $\Rightarrow(\frac{a}{b})^{2 n+1}=1=(\frac{a}{b})^{0}$ $\Rightarrow 2 n+1=0$ $\Rightarrow n=\frac{-1}{2}$Show Answer
Answer : Let the two numbers be $a$ and $b$. G.M. $=\sqrt{a b}$ According to the given condition, $a+b=6 \sqrt{a b}$ $\Rightarrow(a+b)^{2}=36(a b)$ Also, $(a-b)^{2}=(a+b)^{2}-4 a b=36 a b-4 a b=32 a b$ $\Rightarrow a-b=\sqrt{32} \sqrt{a b}$ $=4 \sqrt{2} \sqrt{a b}$ Adding (1) and (2), we obtain $2 a=(6+4 \sqrt{2}) \sqrt{a b}$ $\Rightarrow a=(3+2 \sqrt{2}) \sqrt{a b}$ Substituting the value of $a$ in (1), we obtain $b=6 \sqrt{a b}-(3+2 \sqrt{2}) \sqrt{a b}$ $\Rightarrow b=(3-2 \sqrt{2}) \sqrt{a b}$ $\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$ Thus, the required ratio is $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.Show Answer
prove that the numbers are $A \pm \sqrt{(A+G)(A-G)}$.
Answer : It is given that $A$ and $G$ are A.M. and G.M. between two positive numbers. Let these two positive numbers be $a$ and $b$. $\therefore AM=A=\frac{a+b}{2}$ $GM=G=\sqrt{ab}$ From (1) and (2), we obtain $a+b=2 A$. $a b=G^{2}$. Substituting the value of $a$ and $b$ from (3) and (4) in the identity $(a - b)^{2}=(a+b)^{2}- 4 a b$, we obtain $(a- b)^{2}=4 A^{2}- 4 G^{2}=4(A^{2} -G^{2})$ $(a - b)^{2}=4(A+G)(A$- $G)$ $(a-b)=2 \sqrt{(A+G)(A-G)}$ From (3) and (5), we obtain $
\begin{aligned}
& 2 a=2 A+2 \sqrt{(A+G)(A-G)} \\
& \Rightarrow a=A+\sqrt{(A+G)(A-G)}
\end{aligned}
$ Substituting the value of $a$ in (3), we obtain
$b=2 A-A-\sqrt{(A+G)(A-G)}=A-\sqrt{(A+G)(A-G)}$ Thus, the two numbers are $
A \pm \sqrt{(A+G)(A-G)}
$Show Answer
Answer : It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P. Here, $a=30$ and $r=2$ $\therefore a_3=a r^{2}=(30)(2)^{2}=120$ Therefore, the number of bacteria at the end of $2^{\text{nd }}$ hour will be 120 . $a_5=a r^{4}=(30)(2)^{4}=480$ The number of bacteria at the end of $4^{\text{th }}$ hour will be 480 . $a _{n+1}=a r^{n}=(30) 2^{n}$ Thus, number of bacteria at the end of $n^{\text{th }}$ hour will be $30(2)^{n}$.Show Answer
Answer : The amount deposited in the bank is Rs 500 . At the end of first year, amount $=Rs 500(1+\frac{1}{10})=Rs 500$ (1.1) At the end of $2^{\text{nd }}$ year, amount $=$ Rs 500 (1.1) (1.1) At the end of $3^{\text{rd }}$ year, amount = Rs 500 (1.1) (1.1) (1.1) and so on $\therefore$ Amount at the end of 10 years $=$ Rs 500 (1.1) (1.1) $\ldots$ (10 times) $=Rs 500(1.1)^{10}$Show Answer
Show Answer
Answer :
Let the root of the quadratic equation be $a$ and $b$.
According to the given condition,
A.M. $=\frac{a+b}{2}=8 \Rightarrow a+b=16$
G.M. $=\sqrt{a b}=5 \Rightarrow a b=25$
The quadratic equation is given by,
$x^{2} - x$ (Sum of roots) + (Product of roots) $=0$
$x^{2}- x(a+b)+(a b)=0$
$x^{2} - 16 x+25=0$ [Using (1) and (2)]
Thus, the required quadratic equation is $x^{2}- 16 x+25=0$
Miscellaneous Examples
Example 14 If $a, b, c, d$ and $p$ are different real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2}-2(a b+b c+c d) p+(b^{2}+c^{2}+d^{2}) \leq 0$, then show that $a, b, c$ and $d$ are in G.P.
Solution Given that
$ (a^{2}+b^{2}+c^{2}) p^{2}-2(a b+b c+c d) p+(b^{2}+c^{2}+d^{2}) \leq 0 \quad \quad \quad \quad \quad \ldots (1) $
But L.H.S.
$ =(a^{2} p^{2}-2 a b p+b^{2})+(b^{2} p^{2}-2 b c p+c^{2})+(c^{2} p^{2}-2 c d p+d^{2}), $
which gives $(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2} \geq 0 \quad \quad \quad \quad \quad \ldots (2)$
Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), we have, $\quad(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$
or
$ a p-b=0, b p-c=0, c p-d=0 $
This implies that $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
Hence $a, b, c$ and $d$ are in G.P.
Miscellaneous Exercise On Chapter 8
1. If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{N}$ such that
$ f(1)=3 \text{ and } \sum _{x=1}^{n} f(x)=120 \text{, find the value of } n \text{. } $
Answer : It is given that, $f(x+y)=f(x) \times f(y)$ for all $x, y \in N$. $f(1)=3$ Taking $x=y=1$ in (1), we obtain $f(1+1)=f(2)=f(1) f(1)=3 \times 3=9$ Similarly, $f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$ $f(4)=f(1+3)=f(1) f(3)=3 \times 27=81$ $\therefore f(1), f(2), f(3), \ldots$, that is $3,9,27, \ldots$, forms a G.P. with both the first term and common ratio equal to 3 . It is known that, $
S_n=\frac{a(r^{n}-1)}{r-1}
$ It is given that, $\sum _{x=1}^{n} f(x)=120$
$\therefore 120=\frac{3(3^{n}-1)}{3-1}$ $\Rightarrow 120=\frac{3}{2}(3^{n}-1)$ $\Rightarrow 3^{n}-1=80$ $\Rightarrow 3^{n}=81=3^{4}$ $\therefore n=4$ Thus, the value of $n$ is 4 .Show Answer
Answer : Let the sum of $n$ terms of the G.P. be 315 . It is known that, $S_n=\frac{a(r^{n}-1)}{r-1}$ It is given that the first term $a$ is 5 and common ratio $r$ is 2 . $\therefore 315=\frac{5(2^{n}-1)}{2-1}$ $\Rightarrow 2^{n}-1=63$ $\Rightarrow 2^{n}=64=(2)^{6}$ $\Rightarrow n=6$ $\therefore$ Last term of the G.P $=6^{\text{th }}$ term $=a r^{6 - 1}=(5)(2)^{5}=(5)(32)=160$ Thus, the last term of the G.P. is 160.Show Answer
Answer : Let $a$ and $r$ be the first term and the common ratio of the G.P. respectively. $\therefore a=1$ $a_3=a r^{2}=r^{2}$
$a_5=a r^{4}=r^{4}$ $\therefore r^{2}+r^{4}=90$ $\Rightarrow r^{4}+r^{2}- 90=0$ $\Rightarrow r^{2}=\frac{-1+\sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}=-10$ or 9 $\therefore r= \pm 3$ (Taking real roots) Thus, the common ratio of the G.P. is $\pm 3$.Show Answer
Answer : Let the three numbers in G.P. be $a$, $a r$, and $a r^{2}$. From the given condition, $a+a r+a r^{2}=56$ $\Rightarrow a(1+r+r^{2})=56$ $\Rightarrow a=\frac{56}{1+r+r^{2}}$ a - 1 , ar - $7, a r^{2} - 21$ forms an A.P. $ \therefore (ar - 7)-(a-1) = (ar^2 - 21) - (ar - 7) $ $\Rightarrow ar$ - a - $6=a r^{2} -$ ar - 14 $\Rightarrow a r^{2} - 2 a r+a=8$ $\Rightarrow a r^{2}- a r- a r+a=8$ $\Rightarrow a(r^{2}+1 - 2 r)=8$ $\Rightarrow a(r- 1)^{2}=8$. $\Rightarrow \frac{56}{1+r+r^{2}}(r-1)^{2}=8$ [Using (1)] $\Rightarrow 7(r^{2} - 2 r+1)=1+r+r^{2}$ $\Rightarrow 7 r^{2} - 14 r+7$ - 1 - $r$ - $r^{2}=0$ $\Rightarrow 6 r^{2} - 15 r+6=0$ $\Rightarrow 6 r^{2} - 12 r - 3 r+6=0$ $ \therefore (6ar - 7)-(a-1) = (ar^2 - 21) - (ar - 7) $ $\Rightarrow(6 r- 3)(r$- 2$)=0$
$\therefore r=2, \frac{1}{2}$ When $r=2, a=8$ When $r=\frac{1}{2}, a=32$ Therefore, when $r=2$, the three numbers in G.P. are 8, 16, and 32. When $r=\frac{1}{2}$, the three numbers in G.P. are 32, 16, and 8. Thus, in either case, the three required numbers are 8,16 , and 32 .Show Answer
Answer : Let the G.P. be $T_1, T_2, T_3, T_4, \ldots T _{2 n}$. Number of terms $=2 n$ According to the given condition, $T_1+T_2+T_3+\ldots+T _{2 n}=5[T_1+T_3+\ldots+T _{2_n{-1}}]$ $\Rightarrow T_1+T_2+T_3+ \ldots +T_{2 n}- 5[T_1+T_3+\ldots + T_{2n}] =0$ $\Rightarrow T_2+T_4+\ldots+T _{2 n}=4[T_1+T_3+\ldots+T _{2 n {-1}}]$ Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots$ $\therefore \frac{ar(r^{n}-1)}{r-1}=\frac{4 \times a(r^{n}-1)}{r-1}$ $\Rightarrow a r=4 a$ $\Rightarrow r=4$ Thus, the common ratio of the G.P. is 4.Show Answer
Show Answer
Answer:
It is given that,
$$ \begin{align*} & \frac{a+b x}{a-b x}=\frac{b+c x}{b-c x} \\ & \Rightarrow(a+b x)(b-c x)=(b+c x)(a-b x) \\ & \Rightarrow a b-a c x+b^{2} x-b c x^{2}=a b-b^{2} x+a c x-b c x^{2} \\ & \Rightarrow 2 b^{2} x=2 a c x \\ & \Rightarrow b^{2}=a c \\ & \Rightarrow \frac{b}{a}=\frac{c}{b} \tag{1} \end{align*} $$
Also, $\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$
$\Rightarrow(b+c x)(c-d x)=(b-c x)(c+d x)$
$\Rightarrow b c-b d x+c^{2} x-c d x^{2}=b c+b d x-c^{2} x-c d x^{2}$
$\Rightarrow 2 c^{2} x=2 b d x$
$\Rightarrow c^{2}=b d$
$\Rightarrow \frac{c}{d}=\frac{d}{c}$
From (1) and (2), we obtain
$ \frac{b}{a}=\frac{c}{b}=\frac{d}{c} $
Thus, $a, b, c$, and $d$ are in G.P.
7. Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $P^{2} R^{n}=S^{n}$.
Answer : Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots a r^{n \hat{a} \epsilon^{*}} 1 \ldots$ According to the given information, $
\begin{aligned}
& S=\frac{a(r^{n}-1)}{r-1} \\
& P=a^{n} \times r^{1+2+\ldots+n-1} \\
& =a^{n} r^{\frac{n(n-1)}{2}} \\
& {[\because \text{ Sum of first } n \text{ natural numbers is } n \frac{(n+1)}{2}]} \\
& R=\frac{1}{a}+\frac{1}{a r}+\ldots+\frac{1}{a r^{n-1}} \\
& =\frac{r^{n-1}+r^{n-2}+\ldots r+1}{a r^{n-1}} \\
& =\frac{1(r^{n}-1)}{(r-1)} \times \frac{1}{a r^{n-1}} \quad[\because 1, r, \ldots r^{n-1} \text{ forms a G.P }] \\
& =\frac{r^{n}-1}{a r^{n-1}(r-1)} \\
& \therefore P^{2} R^{n}=a^{2 n} r^{n(n-1)} \frac{(r^{n}-1)^{n}}{a^{n} r^{n(n-1)}(r-1)^{n}} \\
& =\frac{a^{n}(r^{n}-1)^{n}}{(r-1)^{n}} \\
& =[\frac{a(r^{n}-1)}{(r-1)}]^{\prime \prime} \\
& =S^{\prime \prime}
\end{aligned}
$ Hence, $P^{2} R^{n}=S^{n}$Show Answer
Answer : It is given that $a, b, c$, and $d$ are in G.P. $\therefore b^{2}=a c$… $c^{2}=b d$ $a d=b c$ It has to be proved that $(a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P. i.e., $(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$ Consider L.H.S. $(b^{n}+c^{n})^{2}=b^{2 n}+2 b^{n} c^{n}+c^{2 n}$ $=(b^{2})^{n}+2 b^{n} c^{n}+(c^{2})^{n}$ $=(a c)^{n}+2 b^{n} c^{n}+(b d)^{n}[$ Using (1) and (2)] $=a^{n} c^{n}+b^{n} c^{n}+b^{n} c^{n}+b^{n} d^{n}$ $=a^{n} c^{n}+b^{n} c^{n}+a^{n} d^{n}+b^{n} d^{n}$ [Using (3)] $=c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$ $=(a^{n}+b^{n})(c^{n}+d^{n})$ $=$ R.H.S. $\therefore(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$ Thus, $(a^{n}+b^{n}),(b^{n}+c^{n})$, and $(c^{n}+d^{n})$ are in G.P.Show Answer
Answer : It is given that $a$ and $b$ are the roots of $x^{2} \hat{\mathbf{a}} \in$ " $3 x+p=0$ $\therefore a+b=3$ and $a b=p$ Also, $c$ and $d$ are the roots of $x^{2}-12 x+q=0$ $\therefore c+d=12$ and $c d=q$.. It is given that $a, b, c, d$ are in G.P. Let $a=x, b=x r, c=x r^{2}, d=x r^{3}$ From (1) and (2), we obtain $x+x r=3$ $\Rightarrow x(1+r)=3$ $x r^{2}+x r^{3}=12$ $\Rightarrow x r^{2}(1+r)=12$ On dividing, we obtain $\frac{x r^{2}(1+r)}{x(1+r)}=\frac{12}{3}$ $\Rightarrow r^{2}=4$ $\Rightarrow r= \pm 2$ When $r=2, x=\frac{3}{1+2}=\frac{3}{3}=1$ When $r=-2, x=\frac{3}{1-2}=\frac{3}{-1}=-3$ Case I: When $r=2$ and $x=1$, $a b=x^{2} r=2$ $c d=x^{2} r^{5}=32$
$\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$ i.e., $(q+p):(q-p)=17: 15$ Case II: When $r=-2, x=$1, $a b=x^{2} r=-18$ $c d=x^{2} r^{5}=- 288$ $\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}$ i.e., $(q+p):(q-p)=17: 15$ Thus, in both the cases, we obtain $(q+p):(q$ - $p)=17: 15$Show Answer
that $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$.
Answer : Let the two numbers be $a$ and $b$. A.M $=\frac{a+b}{2}$ and G.M. $=\sqrt{a b}$ According to the given condition, $\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}$ $\Rightarrow \frac{(a+b)^{2}}{4(a b)}=\frac{m^{2}}{n^{2}}$ $\Rightarrow(a+b)^{2}=\frac{4 a b m^{2}}{n^{2}}$ $\Rightarrow(a+b)=\frac{2 \sqrt{a b} m}{n}$ Using this in the identity $(a \text{ - } b)^{2}=(a+b)^{2}$ - $4 a b$, we obtain $$
\begin{align*}
& (a-b)^{2}=\frac{4 a b m^{2}}{n^{2}}-4 a b=\frac{4 a b(m^{2}-n^{2})}{n^{2}} \\
& \Rightarrow(a-b)=\frac{2 \sqrt{a b} \sqrt{m^{2}-n^{2}}}{n} \tag{2}
\end{align*}
$$ Adding (1) and (2), we obtain $$
\begin{aligned}
& 2 a=\frac{2 \sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}}) \\
& \Rightarrow a=\frac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})
\end{aligned}
$$ Substituting the value of $a$ in (1), we obtain $$
\begin{aligned}
& b=\frac{2 \sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}}) \\
& =\frac{\sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}} \\
& =\frac{\sqrt{a b}}{n}(m-\sqrt{m^{2}-n^{2}}) \\
& \therefore a: b=\frac{a}{b}=\frac{\frac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})}{\frac{\sqrt{a b}}{n}(m-\sqrt{m^{2}-n^{2}})}=\frac{(m+\sqrt{m^{2}-n^{2}})}{(m-\sqrt{m^{2}-n^{2}})}
\end{aligned}
$$ Thus, $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$ (i) $5+55+555+\ldots$ (ii) $.6+.66+.666+\ldots$ Answer : (i) $5+55+555+\ldots$ Let $S_n=5+55+555+\ldots$. to $n$ terms $
\begin{aligned}
& =\frac{5}{9}[9+99+999+\ldots \text{ to } n \text{ terms }] \\
& =\frac{5}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+\ldots \text{ to } n \text{ terms }] \\
& =\frac{5}{9}[(10+10^{2}+10^{3}+\ldots \text{ n terms })-(1+1+\ldots n \text{ terms })] \\
& =\frac{5}{9}[\frac{10(10^{n}-1)}{10-1}-n] \\
& =\frac{5}{9}[\frac{10(10^{n}-1)}{9}-n] \\
& =\frac{50}{81}(10^{n}-1)-\frac{5 n}{9}
\end{aligned}
$ (ii) $.6+.66+.666+\ldots$ Let $S_n=06 .+0.66+0.666+\ldots$ to $n$ terms $=6[0.1+0.11+0.111+\ldots$ to $n$ terms $]$ $=\frac{6}{9}[0.9+0.99+0.999+\ldots$ to $n$ terms $]$ $=\frac{6}{9}[(1-\frac{1}{10})+(1-\frac{1}{10^{2}})+(1-\frac{1}{10^{3}})+\ldots.$ to n terms $]$ $=\frac{2}{3}[(1+1+\ldots n.$ terms $)-\frac{1}{10}(1+\frac{1}{10}+\frac{1}{10^{2}}+\ldots n.$ terms $.)]$ $=\frac{2}{3}[n-\frac{1}{10}(\frac{1-(\frac{1}{10})^{n}}{1-\frac{1}{10}})]$ $=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}(1-10^{-n})$ $=\frac{2}{3} n-\frac{2}{27}(1-10^{-n})$ Answer : The given series is $2 \times 4+4 \times 6+6 \times 8+\ldots n$ terms $\therefore n^{\text{th }}$ term $=a_n=2 n \times(2 n+2)=4 n^{2}+4 n$ $a _{20}=4(20)^{2}+4(20)=4(400)+80=1600+80=1680$ Thus, the $20^{\text{th }}$ term of the series is 1680 . Answer : It is given that the farmer pays Rs 6000 in cash. Therefore, unpaid amount = Rs 12000 - Rs $6000=$ Rs 6000 According to the given condition, the interest paid annually is $12 %$ of $6000,12 %$ of $5500,12 %$ of $5000, \ldots, 12 %$ of 500 Thus, total interest to be paid $=12 %$ of $6000+12 %$ of $5500+12 %$ of $5000+\ldots+12 %$ of 500 $=12 %$ of $(6000+5500+5000+\ldots+500)$ $=12 %$ of $(500+1000+1500+\ldots+6000)$ Now, the series $500,1000,1500 \ldots 6000$ is an A.P. with both the first term and common difference equal to 500. Let the number of terms of the A.P. be $n$. $\therefore 6000=500+(n$ - 1$) 500$ $\Rightarrow 1+(n- 1)=12$ $\Rightarrow n=12$ $\therefore$ Sum of the A.P $
=\frac{12}{2}[2(500)+(12-1)(500)]=6[1000+5500]=6(6500)=39000
$ Thus, total interest to be paid $=12 %$ of $(500+1000+1500+\ldots+6000)$ $=12 %$ of $39000=$ Rs 4680 Thus, cost of tractor $=($ Rs $12000+$ Rs 4680) $=$ Rs 16680 Answer : It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash. $\therefore$ Unpaid amount $=$ Rs 22000 - Rs $4000=$ Rs 18000 According to the given condition, the interest paid annually is $10 %$ of $18000,10 %$ of $17000,10 %$ of $16000 \ldots 10 %$ of 1000 Thus, total interest to be paid $=10 %$ of $18000+10 %$ of $17000+10 %$ of $16000+\ldots+10 %$ of 1000 $=10 %$ of $(18000+17000+16000+\ldots+1000)$ $=10 %$ of $(1000+2000+3000+\ldots+18000)$ Here, 1000, 2000, $3000 \ldots 18000$ forms an A.P. with first term and common difference both equal to 1000. Let the number of terms be $n$. $\therefore 18000=1000+(n$ - 1 $)(1000)$ $\Rightarrow n=18$ $
\begin{aligned}
& \begin{aligned}
\therefore 1000+2000+\ldots+18000 & =\frac{18}{2}[2(1000)+(18-1)(1000)] \\
& =9[2000+17000] \\
& =171000
\end{aligned} \\
& \begin{aligned}
\therefore \text{ Total interest paid }=10 % \text{ of }(18000+17000+16000+\ldots+1000)
\end{aligned} \\
& =10 % \text{ of Rs } 171000=\text{ Rs } 17100
\end{aligned}
$ Answer : The numbers of letters mailed forms a G.P.: $4,4^{2}, \ldots 4^{8}$ First term $=4$ Common ratio $=4$ Number of terms $=8$ It is known that the sum of $n$ terms of a G.P. is given by $
\begin{aligned}
& S_n=\frac{a(r^{n}-1)}{r-1} \\
& \therefore S_8=\frac{4(4^{8}-1)}{4-1}=\frac{4(65536-1)}{3}=\frac{4(65535)}{3}=4(21845)=87380
\end{aligned}
$ It is given that the cost to mail one letter is 50 paisa. $\therefore$ Cost of mailing 87380 letters $=\operatorname{Rs~} 87380 \times \frac{100}{100}=$ Rs 43690 $
=Rs 87380 \times \frac{50}{100}=\text{ Rs } 43690
$ Thus, the amount spent when $8^{\text{th }}$ set of letter is mailed is Rs 43690 . Answer : It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. $\therefore$ Interest in first year $
=\frac{5}{100} \times Rs 10000=Rs 500
$ $\therefore$ Amount in $15^{\text{th }}$ year $=Rs$ $
10000+\underbrace{500+500+\ldots+500} _{14 \text{ times }}
$ $=$ Rs $10000+14 \times$ Rs 500 = Rs $10000+$ Rs 7000 = Rs 17000 Amount after 20 years $=$ $
Rs 10000+\underbrace{500+500+\ldots+500} _{20 \text{ times }}
$ $=Rs 10000+20 \times Rs 500$ $=Rs 10000+Rs 10000$ = Rs 20000 Answer : Cost of machine $=$ Rs 15625 Machine depreciates by 20% every year. Therefore, its value after every year is $80 %$ of the original cost i.e., $\frac{4}{5}$ of the original cost. $\therefore$ Value at the end of 5 years $=$ $
15625 \times \underbrace{\frac{4}{5} \times \frac{4}{5} \times \ldots \times \frac{4}{5}} _{5 \text{ times }}=5 \times 1024=5120
$ Thus, the value of the machine at the end of 5 years is Rs 5120 . Answer : Let $x$ be the number of days in which 150 workers finish the work. According to the given information, $150 x=150+146+142+\ldots .(x+8)$ terms The series $150+146+142+\ldots .(x+8)$ terms is an A.P. with first term 146 , common difference - 4 and number of terms as $(x+8)$ $
\begin{aligned}
& \Rightarrow 150 x=\frac{(x+8)}{2}[2(150)+(x+8-1)(-4)] \\
& \Rightarrow 150 x=(x+8)[150+(x+7)(-2)] \\
& \Rightarrow 150 x=(x+8)(150-2 x-14) \\
& \Rightarrow 150 x=(x+8)(136-2 x) \\
& \Rightarrow 75 x=(x+8)(68-x) \\
& \Rightarrow 75 x=68 x-x^{2}+544-8 x \\
& \Rightarrow x^{2}+75 x-60 x-544=0 \\
& \Rightarrow x^{2}+15 x-544=0 \\
& \Rightarrow x^{2}+32 x-17 x-544=0 \\
& \Rightarrow x(x+32)-17(x+32)=0 \\
& \Rightarrow(x-17)(x+32)=0 \\
& \Rightarrow x=17 \text{ or } x=-32
\end{aligned}
$ However, $x$ cannot be negative. $\therefore x=17$ Therefore, originally, the number of days in which the work was completed is 17. Thus, required number of days $=(17+8)=25$ By a sequence, we mean an arrangement of number in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type ${1,2,3, \ldots . k}$. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence. Let $a_1, a_2, a_3, \ldots$ be the sequence, then the sum expressed as $a_1+a_2+a_3+\ldots$ is called series. A series is called finite series if it has got finite number of terms. A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by $a$ and its common ratio by $r$. The general or the $n^{\text{th }}$ term of G.P. is given by $a_n=a r^{n-1}$. The sum $S_n$ of the first $n$ terms of G.P. is given by $
S_n=\frac{a(r^{n}-1)}{r-1} \text{ or } \frac{a(1-r^{n})}{1-r} \text{, if } r \neq 1
$ The geometric mean (G.M.) of any two positive numbers $a$ and $b$ is given by $\sqrt{a b}$ i.e., the sequence $a, G, b$ is G.P. Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and geometric sequences. According to Boethius (510), arithmetic and geometric sequences were known to early Greek writers. Among the Indian mathematician, Aryabhatta (476) was the first to give the formula for the sum of squares and cubes of natural numbers in his famous work Aryabhatiyam, written around 499. He also gave the formula for finding the sum to $n$ terms of an arithmetic sequence starting with $p^{\text{th }}$ term. Noted Indian mathematicians Brahmgupta (598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares and cubes. Another specific type of sequence having important applications in mathematics, called Fibonacci sequence, was discovered by Italian mathematician Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification of series into specific forms. In 1671 James Gregory used the term infinite series in connection with infinite sequence. It was only through the rigorous development of algebraic and set theoretic tools that the concepts related to sequence and series could be formulated suitably.Show Answer
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