SEQUENCES AND SERIES - 2 (Geometric Progression)
A sequence of non-zero numbers is called a geometric progression if ratio of a term and the term preceding to it is always a constant. This constant is called the common ratio of the G.P.
i.e. $a _{1}, a _{2}, \ldots \ldots \ldots ., a _{n}$ is in G.P. if $\frac{a _{n+1}}{a _{n}}=r=$ const., $\forall n \varepsilon N$.
$n^{\text {th }}$ term of a G.P. $=a _{n}=a r^{n-1}$.
$n$th term from end of a G.P $=a _{n} \cdot\left(\frac{1}{r}\right)^{n-1}$ where $a _{n}$ is the last term and $r$ the common ratio of the G.P.
| Sign of a | $\mathbf{+ v e}$ | $\mathbf{+ v e}$ | $\mathbf{- v e}$ | $-\mathbf{v e}$ |
|---|---|---|---|---|
| Range of $\mathrm{r}$ | $\mathrm{r}>1$ | $0<\mathbf{r}<1$ | $\mathrm{r}>1$ | $0<\mathbf{r}<1$ |
| G.P. is | increasing | decreasing | decreasing | increasing |
If $\mathrm{r}=1$, the sequence will be a constant sequence. If $\mathrm{r}$ is negative the terms of G.P. are alternately positive and negative and so the G.P is neither increasing nor decreasing.
Sum of $\mathbf{n}$ terms of a G.P
Let the first term of G.P be a common ratio $r$ and last term $a _{n}$, then
i. Sum to $n$ terms
$ \mathrm{S} _{\mathrm{n}}=\left\{\begin{aligned} \frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1} \text { or } \mathrm{a}\left(\frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}\right) & ; \text { when } \mathrm{r} \neq 1 \\ \mathrm{na} & ; \text { when } \mathrm{r}=1 \end{aligned}\right. $
Also $S _{n}=\frac{a-a _{n} r}{1-r}$ or $\frac{a _{n} r-a}{r-1} ; \quad$ when $\mathrm{r} \neq 1$
Sum of an Infinite G.P
Sum of an infinite G.P. with first term a and common ratio $\mathrm{r}(-1<\mathrm{r}<1 ; \mathrm{r} \neq 0$ or $0<|\mathrm{r}|<1)$ is given by $\mathrm{S}=\frac{\mathrm{a}}{1-\mathrm{r}}$
If $r \geq 1$, then the sum of an infinite G.P tends to infinity.
Selection of terms in a G.P
In case of an odd number of terms the middle term is a and common ratio is $r$ while in case of even number of terms, middle terms are $\frac{\mathrm{a}}{\mathrm{r}}$, ar and common ratio is $\mathrm{r}^{2}$.
| No. of terms | Terms | Common ratio |
|---|---|---|
| 3 | $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar}$ | $\mathrm{r}$ |
| 4 | $\frac{\mathrm{a}}{\mathrm{r}^{3}}, \frac{\mathrm{a}}{\mathrm{r}}, \mathrm{ar}, \mathrm{ar}^{3}$ | $\mathrm{r}^{2}$ |
| 5 | $\frac{\mathrm{a}}{\mathrm{r}^{2}}, \frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}$ | $\mathrm{r}$ |
Note that it is convenient to take the terms as a, ar, $\mathrm{ar}^{2}$ ………… if the product of numbers is not given.
Insertion of geometric means
Note : If $\mathrm{a} \& \mathrm{~b}$ are two numbers of opposite signs then geometric mean between them does not exist.
Solved examples
1. Suppose $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{A} . P$ and $\mathrm{a}^{2}, \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in G.P. If $\mathrm{a}<\mathrm{b}<\mathrm{c}$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{3}{2}$, then the value of a is
(a). $\frac{1}{2 \sqrt{2}}$
(b). $\frac{1}{2 \sqrt{3}}$
(c). $1 \frac{1}{2}-\frac{1}{\sqrt{3}}$
(d). $\frac{1}{2}-\frac{1}{\sqrt{2}}$
Show Answer
Solution: Let the numbers be $\mathrm{A}-\mathrm{d}, \mathrm{A}, \mathrm{A}+\mathrm{d}$
Then $\mathrm{A}=\frac{1}{2}$
$\therefore \quad$ Numbers are $\frac{1}{2}-\mathrm{d}, \frac{1}{2}, \frac{1}{2}+\mathrm{d}$
$\mathrm{a}^{2}, \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in G.P.
$\left(\left(\frac{1}{2}\right)^{2}\right)^{2}=\left(\frac{1}{2}-\mathrm{d}\right)^{2}\left(\frac{1}{2}+\mathrm{d}\right)^{2}$
$\Rightarrow \quad \frac{1}{16}=\left(\frac{1}{4}-\mathrm{d}^{2}\right)^{2} \Rightarrow \frac{1}{4}-\mathrm{d}^{2}= \pm \frac{1}{4}$
$\Rightarrow \quad \mathrm{d}^{2}=\frac{1}{2} \quad \Rightarrow \mathrm{d}= \pm \frac{1}{\sqrt{2}}$
Since $\mathrm{a}<\mathrm{b}<\mathrm{c}, \mathrm{a}=\frac{1}{2}-\frac{1}{\sqrt{2}}$
Answer: d
2. If $\mathrm{a} _{\mathrm{n}}=\frac{3}{4}-\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}+\ldots \ldots \ldots+(-1)^{\mathrm{n}-1}\left(\frac{3}{4}\right)^{\mathrm{n}}$ and $\mathrm{b} _{\mathrm{n}}=1-\mathrm{a} _{\mathrm{n}}$, then the minimum natural number $n _{0}$ such that $b _{n}>a _{n} \& n>n _{0}$ is
(a). 4
(b). 5
(c). 6
(d). 12
Show Answer
Solution: $\ \mathrm{a} _{\mathrm{n}}=\frac{\frac{3}{4}\left(1-\left(\frac{-3}{4}\right)^{\mathrm{n}}\right)}{1+\frac{3}{4}}$
i.e. $a _{n}=\frac{3}{7}\left(1-\left(\frac{-3}{4}\right)^{n}\right)$
$b _{n}>a _{n} \Rightarrow 1-a _{n}>a _{n} \Rightarrow 2 a _{n}<1$
$\Rightarrow \quad \frac{6}{7}\left(1-\left(\frac{-3}{4}\right)^{\mathrm{n}}\right)<1 \quad \Rightarrow \quad 1-\left(\frac{-3}{4}\right)^{\mathrm{n}}<\frac{7}{6} \quad-\left(\frac{-3}{4}\right)^{\mathrm{n}}<\frac{1}{6}$
$\Rightarrow \quad(-3)^{n+1}<2^{2 n-1}$
For $\mathrm{n}$ to be even, the inequality always holds. For $\mathrm{n}$ to be odd, it holds for $\mathrm{n} \geq 7$.
$\therefore \quad$ Least natural number for which it holds is 6 .
Answer: c
3. If $a$ be the arithmetic mean of $b$ and $c$ and $\mathrm{G} _{1}, \mathrm{G} _{2}$ be the two geometric means between them, then $\mathrm{G} _{1}{ }^{3}+\mathrm{G} _{2}{ }^{3}=$
(a). $\mathrm{G} _{1} \mathrm{G} _{2} \mathrm{a}$
(b). $2 \mathrm{G} _{1} \mathrm{G} _{2} \mathrm{a}$
(c). $3 \mathrm{G} _{1} \mathrm{G} _{2} \mathrm{a}$
(d). none of these
Show Answer
Solution: $b, a, c$ are in A.P
$2 \mathrm{a}=\mathrm{b}+\mathrm{c}$
Also $\mathrm{b}, \mathrm{G} _{1}, \mathrm{G} _{2}, \mathrm{c}$ are in G.P
$\mathrm{G} _{1}{ }^{2}=\mathrm{bG} _{2} \quad \Rightarrow \quad \mathrm{G} _{1}{ }^{3}=\mathrm{bG} _{1} \mathrm{G} _{2}$
$\mathrm{G} _{2}{ }^{2}=\mathrm{G} _{1} \mathrm{c} \quad \Rightarrow \quad \mathrm{G} _{2}{ }^{3}=\mathrm{cG} _{1} \mathrm{G} _{2}$
$\mathrm{G} _{1}{ }^{3}+\mathrm{G} _{2}{ }^{3}=\mathrm{G} _{1} \mathrm{G} _{2}(\mathrm{~b}+\mathrm{c})=2 \mathrm{G} _{1} \mathrm{G} _{2} \mathrm{a}$
Answer: b
4. It is known that $\sum _{\mathrm{r}=1}^{\infty} \frac{1}{(2 \mathrm{r}-1)^{2}}=\frac{\pi^{2}}{8}$, then $\sum _{\mathrm{r}=1}^{\infty} \frac{1}{\mathrm{r}^{2}}$ is
(a). $\frac{\pi^{2}}{24}$
(b). $\frac{\pi^{2}}{3}$
(c). $\frac{\pi^{2}}{6}$
(d). none of these
Show Answer
Solution:
Let $\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots \ldots \ldots \infty=\mathrm{x}$
$\sum _{\mathrm{r}=1}^{\infty} \frac{1}{\mathrm{r}^{2}}=\left(\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\ldots \ldots \ldots ..\right)+\left(\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+\ldots \ldots \ldots . . . \infty\right)$
$\mathrm{x}=\frac{\pi^{2}}{8}+\frac{1}{4}\left(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots \ldots \ldots \infty\right)$
$x=\frac{\pi^{2}}{8}+\frac{1}{4} x \quad \Rightarrow \quad \frac{3 x}{4}=\frac{\pi^{2}}{8} \quad \Rightarrow \quad x=\frac{\pi^{2}}{6}$
Answer: c
5. If $\mathrm{a} _{\mathrm{i}} \varepsilon \mathrm{R}, \mathrm{i}=1,2,3, \ldots \ldots . \mathrm{n}$ and all $\mathrm{a} _{\mathrm{i}}{ }^{\prime} \mathrm{s}$ are distinct such that $\left(\sum _{i=1}^{n-1} a _{i}{ }^{2}\right) x^{2}+2\left(\sum _{i=1}^{n-1} a _{i} a _{i+1}\right) x+\sum _{i=2}^{n} a _{i}^{2} \leq 0$, then $a _{1}, a _{2}, \ldots \ldots . .$. are in
(a). G.P
(b). A.P
(c). H.P
(d). none of these
Show Answer
Solution: $\quad \sum _{i=1}^{n-1}\left(a _{i} x+a _{i+1}\right)^{2} \leq 0 \quad \Rightarrow \quad \sum _{i=1}^{n-1}\left(a _{i} x+a _{i+1}\right)^{2}=0$
$ \begin{aligned} & \Rightarrow \quad a _{i} x+a _{i+1}=0 \quad \forall \mathrm{i}=1,2, \ldots \ldots \ldots n-1 \\ & \Rightarrow \quad \frac{a _{i+1}}{a _{i}}=-x \forall i=1,2, \ldots \ldots . n-1 \\ & \therefore \quad a _{1}, a _{2}, \ldots \ldots \ldots a _{n} \text { are in G.P. } \end{aligned} $
Answer: a
6. The $1025^{\text {th }}$ term is the sequence
$1, 22, 4444, 88888888, \ldots \ldots \ldots .$. is
(a). $2^{9}$
(b). $2^{10}$
(c). $2^{11}$
(d). none of these
Show Answer
Solution: Number of digits in each term are in G.P.
Let 1025 th term $=2^{\mathrm{n}}$
then
$ \begin{aligned} & 1+2+4+8+\ldots \ldots .+2^{n-1}<1025 \leq 1+2+4+8+\ldots . .+2^{n} \\ & \Rightarrow \quad 1 \frac{\left(2^{n}-1\right)}{2-1}<1025 \leq 1 \cdot \frac{\left(2^{n+1}-1\right)}{2-1} \\ & \Rightarrow \quad 2^{n}-1<1025 \leq 2^{n+1}-1 \\ & \Rightarrow \quad 2^{n+1} \geq 1026>1024=2^{10} \\ & \Rightarrow \quad \mathrm{n}+1>10 \Rightarrow \mathrm{n}>9 \\ & \therefore \quad \mathrm{n}=10 \end{aligned} $
Answer: b
7. If $\mathrm{a}^{1 / \mathrm{x}}=\mathrm{b}^{1 / \mathrm{y}}=\mathrm{c}^{1 / \mathrm{z}}$ and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in geometrical progression, then $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in
(a). A.P.
(b). G.P.
(c). H.P.
(d). None of these
Show Answer
Solution: $\mathrm{a}^{1 / \mathrm{x}}=\mathrm{b}^{1 / \mathrm{y}}=\mathrm{c}^{1 / \mathrm{z}}=\mathrm{k}$
$\Rightarrow \quad \mathrm{a}=\mathrm{k}^{\mathrm{x}}, \mathrm{b}=\mathrm{k}^{\mathrm{y}}, \mathrm{c}=\mathrm{k}^{\mathrm{z}}$
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P. $\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
$\Rightarrow \mathrm{k}^{2 \mathrm{y}}=\mathrm{k}^{\mathrm{x}+\mathrm{z}}$
$\Rightarrow 2 \mathrm{y}=\mathrm{x}+\mathrm{z}$
Answer: a
8. If the arithmetic mean of two numbers be A and geometric mean be G, then the numbers will be
(a). $\mathrm{A} \pm \sqrt{\left(\mathrm{A}^{2}-\mathrm{G}^{2}\right)}$
(b). $\sqrt{\mathrm{A}} \pm \sqrt{\mathrm{A}^{2}-\mathrm{G}^{2}}$
(c). $\pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$
(d).$\frac{A \pm \sqrt{(A+G)(A-G)}}{2}$
Show Answer
Solution: Let the number be $a$ and $b$
$A=\frac{a+b}{2} \quad G^{2}=a b$
$a$ and $b$ are the roots of $x^{2}-2 A x+G^{2}=0$
$ \begin{array}{r} \mathrm{x}=\frac{2 \mathrm{~A} \pm \sqrt{4 \mathrm{~A}^{2}-4 \mathrm{G}^{2}}}{2} \\ \mathrm{x}=\mathrm{A} \pm \sqrt{\left(\mathrm{A}^{2}-\mathrm{G}^{2}\right)} \end{array} $
Answer: a
Practice questions
1. The sum of an infinite geometric series is 2 and the sum of the geometric series made from the cubes of this infinite series is 24 . Then the series is
(a). $3+\frac{3}{2}-\frac{3}{4}+\frac{3}{8} \ldots \ldots$
(b). $3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8} \ldots$.
(c). $3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8} \ldots .$.
(d). none of these
Show Answer
Answer: (c)2. Read the passage and answer the following questions.
Let $A _{1}, A _{2}, \ldots . . A _{m}$ be arithmetic means between -2 and 1027 and $G _{1}, G _{2}, \ldots . \mathrm{G} _{\mathrm{n}}$ be Geometric means between 1 and 1024. Product of geometric means is $2^{45}$ and sum of arithmetic means is $1025 \times 171$.
i. The value of $n$ is
(a). 7
(b). 9
(c). 11
(d). none of these
Show Answer
Answer: (b)ii. The value of $m$ is
(a). 340
(b). $ 342$
(c). 344
(d). 346
Show Answer
Answer: (b)iii. The value of $\mathrm{G} _{1}+\mathrm{G} _{2}+\mathrm{G} _{3}+\ldots .+\mathrm{G} _{\mathrm{n}}$ is
(a). 1022
(b). 2044
(c). 512
(d). none of these
Show Answer
Answer: (a)iv. The common difference of the progression $\mathrm{A} _{1}, \mathrm{~A} _{3}, \mathrm{~A} _{5} \ldots . . \mathrm{An}$ is
(a). 6
(b). 3
(c). 2
(d). 1
Show Answer
Answer: (a)v. The numbers $2 \mathrm{~A} _{171}, \mathrm{G} _{5}^{2}+1,2 \mathrm{~A} _{172}$ are in
(a). $\mathrm{AP}$
(b). GP
(c). $\mathrm{HP}$
(d). AGP
Show Answer
Answer: (a)3. The difference between two numbers is 48 and the difference between their arithmetic mean and their geometric mean is 18 . Then, the greater of two numbers is
(a). 96
(b). 60
(c). 54
(d). 49
Show Answer
Answer: (d)4. If $\mathrm{a}^{\mathrm{x}}=\mathrm{b}^{\mathrm{y}}=\mathrm{c}^{\mathrm{z}}=\mathrm{d}^{\mathrm{w}}$, the value of $\mathrm{x}\left(\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+\frac{1}{\mathrm{w}}\right)$ is
(a). $\log _{a}(a b c)$
(b). $ \log _{2}(\mathrm{bcd})$
(c). $\log _{\mathrm{b}}(\mathrm{cda})$
(d). $\log _{c}(\mathrm{dab})$
Show Answer
Answer: (b)5. If three positive numbers $x, y, z$ are in A.P. and also $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P., then
(a). $\mathrm{x}=\mathrm{y}=\mathrm{z}$
(b). $ x \neq y=z$
(c). $ x=y \neq z$
(d). none of these
Show Answer
Answer: (a)6. If $a _{1}, a _{2}, a _{3}$ are three consecutive terms of a G.P. with common ratio $k$. Then the values of $k$ for which the inequality $a _{3}>4 a _{2}-3 a _{1}$, is satisfied is (if $a _{1}>0$ )
(a). $(1,3)$
(b). $(-\infty, 1) \cup(3, \infty)$
(c). $\mathrm{R}$
(d). none of these
Show Answer
Answer: (b)7. The three successive terms of a G.P. will form the side of a triangle if the common ratio $r$ lies in
(a). $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(b). $\left(\frac{-1 \sqrt{5}}{2}, \infty\right)$
(c). $\left(-\infty, \frac{\sqrt{5}-1}{2}\right)$
(d). none of these
Show Answer
Answer: (a)8. If for $\left.0<x<\frac{\pi}{2}, \exp \left(\sin ^{2} x+\sin ^{4} x+\sin ^{6} x+\ldots \ldots \infty\right) \log _{e} 2\right)$ satisfies the quadratic equation $\mathrm{x}^{2}-9 \mathrm{x}+8=0, \frac{\sin \mathrm{x}-\cos \mathrm{x}}{\sin \mathrm{x}+\cos \mathrm{x}}$ is
(a). $2-\sqrt{3}$
(b). $2+\sqrt{3}$
(c). $\sqrt{3}-2$
(d). none of these
Show Answer
Answer: (a)9. Match the following
| Column I | Column II |
|---|---|
| (a). If $a, b, c$ are non zero real numbers such that $3\left(a^{2}+b^{2}+c^{2}+1\right)=2(a+b+c+a b+b c+c a)$ then $a, b, c$ are in |
(p) AP |
| (b). If the square of difference of three numbers be in AP, then their differences are in | (q) GP |
| (c). If $a-b, a x-b y, a x^{2}-b y^{2}(a, b \neq 0)$ are in G.P. then $\mathrm{x}, \mathrm{y}, \frac{\mathrm{ax}-\mathrm{by}}{\mathrm{a}-\mathrm{b}}$ are in | (r) HP |
| (s) equal |
Show Answer
Answer: a $\rarr$ p, q, s; b $\rarr$ r; c $\rarr$ p, q, s10. If $1+p+p^{2}+\ldots \ldots+p^{n}=(1+p)\left(1+p^{2}\right)\left(1+p^{4}\right)\left(1+p^{8}\right)\left(1+p^{16}\right)$ then the value of $n(n \varepsilon N)$ is
(a). 32
(b). 16
(c). 31
(d). 15
Show Answer
Answer: (c)11. If $\sin \theta, \sqrt{2}(\sin \theta+1), 6 \sin \theta+6$ are in G.P., then the fifth term is
(a). $81$
(b). $82 \sqrt{2}$
(c). $162$
(d). none of these
Show Answer
Answer: (c)12. If $\mathrm{x} \varepsilon\{1,2,3, \ldots .9\}$ and $\mathrm{f} _{\mathrm{n}}(\mathrm{x})=\mathrm{xxx} \ldots \ldots \mathrm{x}\left(\mathrm{n}\right.$ digits) then $\mathrm{f} _{\mathrm{n}}^{2}(3)+\mathrm{f} _{\mathrm{n}}(2)$ is equal to
(a). $2 \mathrm{f} _{2 \mathrm{n}}(1)$
(b). $\mathrm{f} _{\mathrm{n}}^{2}(1)$
(c). $f _{2 n}(1)$
(d). $2 \mathrm{f} _{2 \mathrm{n}}(4)$
Show Answer
Answer: (c)13. The number of divisors of 1029,1547 and 122 are in
(a). $\mathrm{AP}$
(b). GP
(c). $\mathrm{HP}$
(d). none of these
Show Answer
Answer: (a)14. Let $\mathrm{x} _{1}, \mathrm{x} _{2}, \ldots ., \mathrm{x} _{n}$ be a sequence of integers such that
(i) $-1 \leq \mathrm{x} _{\mathrm{i}} \leq 2$ for $\mathrm{i}=1,2, \ldots \mathrm{n}$
(ii) $\mathrm{x} _{\mathrm{i}}+\mathrm{x} _{2}+\ldots+\mathrm{x} _{\mathrm{n}}=19$
(iii) $\mathrm{x} _{1}{ }^{2}+\mathrm{x} _{2}{ }^{2}+\ldots+\mathrm{x} _{\mathrm{n}}{ }^{2}=99$
Let $\mathrm{m}$ and $\mathrm{M}$ be the minimum and Maximum possible values of $\mathrm{x} _{1}{ }^{3}+\mathrm{x} _{2}{ }^{3}+\ldots . \mathrm{x} _{\mathrm{n}}{ }^{3}$ respectively, then the value of $\frac{M}{m}$ is……..
Show Answer
Answer: 715. Let $16,4,1, \ldots$ be a geometric sequence. Define $P _{n}$ as the product of the first $n$ terms. Then the value of $\frac{\sum _{n=1}^{\infty} \sqrt[n]{P _{n}}}{4}$ is……..
(a). $64$
(b). $\frac{1}{64}$
(c). $32$
(d). none of these
