# NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.3

## Chapter 9 Ex.9.3 Question 1

Find the $${20^{th}}$$ and $${n^{th}}$$ terms of the G.P. $$\frac{5}{2},\frac{5}{4},\frac{5}{8}, \ldots$$

### Solution

The given G.P is $$\frac{5}{2},\frac{5}{4},\frac{5}{8},................$$

Here,$$a = \frac{5}{2}$$ and \begin{align}r = \frac{{{\raise0.7ex\hbox{5} \!\mathord{\left/{\vphantom {5 }}\right.}\!\lower0.7ex\hbox{4}}}}{{{\raise0.7ex\hbox{5} \!\mathord{\left/{\vphantom {5 2}}\right.}\!\lower0.7ex\hbox{2}}}} = \frac{1}{2}\end{align}

Therefore,

\begin{align}{a_{20}}&= a{r^{20 - 1}} = \frac{5}{2}{\left( {\frac{1}{2}} \right)^{19}} = \frac{5}{{\left( 2 \right){{\left( 2 \right)}^{19}}}} = \frac{5}{{{{\left( 2 \right)}^{20}}}}\\{a_n}&= a{r^{n - 1}} = \frac{5}{2}{\left( {\frac{1}{2}} \right)^{n - 1}} = \frac{5}{{\left( 2 \right){{\left( 2 \right)}^{n - 1}}}} = \frac{5}{{{{\left( 2 \right)}^n}}}\end{align}

## Chapter 9 Ex.9.3 Question 2

Find the $${12^{th}}$$ term of a G.P whose $${8^{th}}$$ term is 192 and the common ratio is 2.

### Solution

Let $$a$$ be the first term of the G.P

It is given that common ratio, $$r = 2$$

Eighth term of the G.P, $${a_8} = 192$$

Therefore,

\begin{align}&\Rightarrow {a_8} = a{r^{8 - 1}} = a{r^7}\\&\Rightarrow a{r^7} = 192\\&\Rightarrow a{\left( 2 \right)^7} = 192\\&\Rightarrow a{\left( 2 \right)^7} = {\left( 2 \right)^6}\left( 3 \right)\\&\Rightarrow a = \frac{{{{\left( 2 \right)}^6}\left( 3 \right)}}{{{{\left( 2 \right)}^7}}} = \frac{3}{2}\end{align}

Hence,

\begin{align}{a_{12}} &= a{r^{12 - 1}}\\&= a{r^{11}}\\&= \left( {\frac{3}{2}} \right){\left( 2 \right)^{11}}\\&= \left( 3 \right){\left( 2 \right)^{10}}\\&= 3072\end{align}

## Chapter 9 Ex.9.3 Question 3

The $${5^{th}},\;{8^{th}}$$and $${11^{th}}$$ terms of a G.P are $$p,\;q$$ and $$s$$ respectively. Show that $${q^2} = ps$$ .

### Solution

Let $$a$$ be the first term and $$r$$ be the common ratio of the G.P.

According to the question,

\begin{align}{a_5}&= a{r^{5 - 1}} = a{r^4} = p\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{a_8}&= a{r^{8 - 1}} = a{r^7} = q\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\{a_{11}}&= a{r^{11 - 1}} = a{r^{10}} = s\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Dividing $$\left( 2 \right)$$ by $$\left( 1 \right)$$, we obtain

\begin{align}\frac{{a{r^7}}}{{a{r^4}}}&= \frac{q}{p}\\{r^3}&= \frac{q}{p}\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

Dividing $$\left( 3 \right)$$ by $$\left( 2 \right)$$, we obtain

\begin{align}\frac{{a{r^{10}}}}{{a{r^7}}}&= \frac{s}{q}\\{r^3}&= \frac{s}{q}\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\end{align}

Equating the values of $${r^3}$$ obtained in $$\left( 4 \right)$$ and $$\left( 5 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{q}{p} = \frac{s}{q}\\&\Rightarrow {q^2} = ps\end{align}

Hence proved.

## Chapter 9 Ex.9.3 Question 4

The $${4^{th}}$$term of a G.P is square of its second term, and the first term is $$- 3$$. Determine its $${7^{th}}$$ term.

### Solution

Let be the first term and $$r$$ be the common ratio of the G.P.

It is known that $${{a}_{n}}=a{{r}^{n-1}}$$

Therefore,

\begin{align} & {{a}_{2}}=a{{r}^{1}}=\left( -3 \right)r \\ & {{a}_{4}}=a{{r}^{3}}=\left( -3 \right){{r}^{3}} \\ \end{align}

According to the question,

\begin{align}&\Rightarrow \left( { - 3} \right){r^3} = {\left[ {\left( { - 3} \right)r} \right]^2}\\&\Rightarrow - 3{r^3} = 9{r^2}\\&\Rightarrow r = - 3\end{align}

Hence,

\begin{align}{a_7}&= a{r^{7 - 1}}\\&= a{r^6}\\&= \left( { - 3} \right){\left( { - 3} \right)^6}\\&= {\left( { - 3} \right)^7}\\&= - 2187\end{align}

Thus, the seventh term of the G.P is $$- 2187$$.

## Chapter 9 Ex.9.3 Question 5

Which term of the following sequences:

$$2,2\sqrt 2 ,4 \ldots$$ is $$128$$?

$$\sqrt 3 ,3,3\sqrt 3 , \ldots$$ is $$729$$?

$$\frac{1}{3},\frac{1}{9},\frac{1}{{27}}, \ldots$$ is $$\frac{1}{{19683}}$$?

### Solution

The given sequence is $$2,2\sqrt 2 ,4 \ldots$$

Here, $$a = 2$$ and $$r = \frac{{2\sqrt 2 }}{2} = \sqrt 2$$

Let the $${n^{th}}$$ term of the sequence be 128

\begin{align}&\Rightarrow {a_n} = a{r^{n - 1}} = 128\\&\Rightarrow \left( 2 \right){\left( {\sqrt 2 } \right)^{n - 1}} = 128\\&\Rightarrow \left( 2 \right){\left( 2 \right)^{\frac{{n - 1}}{2}}} = {\left( 2 \right)^7}\\&\Rightarrow {\left( 2 \right)^{\frac{{n - 1}}{2} + 1}} = {\left( 2 \right)^7}\end{align}

Hence,

\begin{align}&\Rightarrow \frac{{n - 1}}{2} + 1 = 7\\&\Rightarrow \frac{{n - 1}}{2} = 6\\&\Rightarrow n - 1 = 12\\&\Rightarrow n = 13\end{align}

Thus, the $${13^{th}}$$ term of the sequence be 128.

The given sequence is $$\sqrt 3 ,3,3\sqrt 3 , \ldots$$

Here, $$a = \sqrt 3$$ and $$r = \frac{3}{{\sqrt 3 }} = \sqrt 3$$

Let the $${n^{th}}$$term of the sequence be $$729$$.

\begin{align}&\Rightarrow {a_n} = a{r^{n - 1}} = 729\\&\Rightarrow \left( {\sqrt 3 } \right){\left( {\sqrt 3 } \right)^{n - 1}} = 729\\&\Rightarrow {\left( 3 \right)^{\frac{1}{2}}}{\left( 3 \right)^{\frac{{n - 1}}{2}}} = {\left( 3 \right)^6}\\&\Rightarrow {\left( 3 \right)^{\frac{1}{2} + \frac{{n - 1}}{2}}} = {\left( 3 \right)^6}\end{align}

Hence,

\begin{align}&\Rightarrow \frac{1}{2} + \frac{{n - 1}}{2} = 6\\&\Rightarrow \frac{{1 + n - 1}}{2} = 6\\&\Rightarrow n = 12\end{align}

Thus, the $${12^{th}}$$ term of the sequence is $$729$$.

The given sequence is $$\frac{1}{3},\frac{1}{9},\frac{1}{{27}}, \ldots$$

Here, $$a = \frac{1}{3}$$ and \begin{align}r = \frac{{{\raise0.7ex\hbox{1} \!\mathord{\left/{\vphantom {9 }}\right.}\!\lower0.7ex\hbox{9}}}}{{{\raise0.7ex\hbox{1} \!\mathord{\left/{\vphantom {3 2}}\right.}\!\lower0.7ex\hbox{3}}}} = \frac{1}{3}\end{align}

Let the $${n^{th}}$$ term of the sequence be $$\frac{1}{{19683}}$$.

\begin{align}&\Rightarrow {a_n} = a{r^{n - 1}} = \frac{1}{{19683}}\\&\Rightarrow \left( {\frac{1}{3}} \right){\left( {\frac{1}{3}} \right)^{n - 1}} = \frac{1}{{19683}}\\&\Rightarrow {\left( {\frac{1}{3}} \right)^n} = {\left( {\frac{1}{3}} \right)^9}\\&\Rightarrow n = 9\end{align}

Thus, the $${9^{th}}$$ term of the sequence be $$\frac{1}{{19683}}$$.

## Chapter 9 Ex.9.3 Question 6

For what values of $$x,$$ the numbers $$- \frac{2}{7},x, - \frac{7}{2}$$ are in G.P?

### Solution

The given numbers are $$- \frac{2}{7},x, - \frac{7}{2}$$

Hence,

Common ratio \begin{align}r = \frac{x}{{{\raise0.7ex\hbox{-2} \!\mathord{\left/{\vphantom { 2}}\right.}\!\lower0.7ex\hbox{7}}}} =- \frac{7x}{2}\end{align}

Also, Common ratio\begin{align}r = \frac{{{\raise0.7ex\hbox{-7} \!\mathord{\left/{\vphantom {2 }}\right.}\!\lower0.7ex\hbox{2}}}}{x} = \frac{7}{2x}\end{align}

Therefore,

\begin{align}&\Rightarrow - \frac{{7x}}{2} = - \frac{7}{{2x}}\\&\Rightarrow 14{x^2} = 14\\&\Rightarrow {x^2} = 1\\&\Rightarrow x = \pm \sqrt 1 \\&\Rightarrow x = \pm 1\end{align}

Thus, for$$x = \pm 1$$, the given numbers will be in G.P.

## Chapter 9 Ex.9.3 Question 7

Find the sum up to $$20$$ terms in the G.P $$0.15,0.015,0.0015 \ldots$$

### Solution

The given G.P is $$0.15,0.015,0.0015 \ldots$$

Here, $$a = 0.15$$ and $$r = \frac{{0.015}}{{0.15}} = 0.1$$

It is known that $${S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$$

Therefore,

\begin{align}{S_{20}} &= \frac{{0.15\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]}}{{1 - 0.1}}\\&= \frac{{0.15}}{{0.9}}\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]\\&= \frac{1}{6}\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]\end{align}

## Chapter 9 Ex.9.3 Question 8

Find the sum of $$n$$ terms in the G.P $$\sqrt 7 ,\sqrt {21} ,3\sqrt 7 , \ldots$$

### Solution

The given G.P is $$\sqrt 7 ,\sqrt {21} ,3\sqrt a \ne - 17 , \ldots$$

Here$$a = \sqrt 7$$, and $$r = \frac{{\sqrt {21} }}{{\sqrt 7 }} = \sqrt 3$$

It is known that $${S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$$

Therefore,

\begin{align}{S_n}&= \frac{{\sqrt 7 \left[ {1 - {{\left( {\sqrt 3 } \right)}^n}} \right]}}{{1 - \sqrt 3 }}\\&= \frac{{\sqrt 7 \left[ {1 - {{\left( {\sqrt 3 } \right)}^n}} \right]}}{{1 - \sqrt 3 }} \times \frac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}\\&= \frac{{\sqrt 7 \left( {\sqrt 3 + 1} \right)\left[ {1 - {{\left( {\sqrt 3 } \right)}^n}} \right]}}{{1 - 3}}\\&= \frac{{ - \sqrt 7 \left( {\sqrt 3 + 1} \right)\left[ {1 - {{\left( {\sqrt 3 } \right)}^n}} \right]}}{2}\\&= \frac{{\sqrt 7 \left( {\sqrt 3 + 1} \right)}}{2}\left[ {{{\left( 3 \right)}^{\frac{n}{2}}} - 1} \right]\end{align}

## Chapter 9 Ex.9.3 Question 9

Find the sum of $$n$$ terms in the G.P $$1, - a,{a^2}, - {a^3}, \ldots$$ (if ).

### Solution

The given G.P is $$1, - a,{a^2}, - {a^3}, \ldots$$

Here, $${a_1} = 1$$ and $$r = - a$$

It is known that $${S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$$

Therefore,

\begin{align}{S_n} &= \frac{{1\left[ {1 - {{\left( { - a} \right)}^n}} \right]}}{{1 - \left( { - a} \right)}}\\&= \frac{{\left[ {1 - {{\left( { - a} \right)}^n}} n\right]}}{{1 + a}}\end{align}

## Chapter 9 Ex.9.3 Question 10

Find the sum of terms in the G.P. $${x^3},{x^5},{x^7}, \ldots$$ (if $$x \ne \pm 1$$).

### Solution

The given G.P is $${x^3},{x^5},{x^7}, \ldots$$

Here, $$a = {x^3}$$ and $$r = {x^2}$$

It is known that $${S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$$

Therefore,

\begin{align}{{S}_{n}} & =\frac{{{x}^{3}}\left[ 1-{{\left( {{x}^{2}} \right)}^{n}} \right]}{1-{{x}^{2}}} \\ & =\frac{{{x}^{3}}\left( 1-{{x}^{2n}} \right)}{1-{{x}^{2}}} \end{align}

## Chapter 9 Ex.9.3 Question 11

Evaluate $$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)}$$

### Solution

\begin{align}\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)}&= \sum\limits_{k = 1}^{11} {\left( 2 \right) + } \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right)} \\&= 22 + \sum\limits_{k = 1}^{11} {{3^k}\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)} \end{align}

Now,

$\sum\limits_{k = 1}^{11} {{3^k}} = {3^1} + {3^2} + {3^3} + \ldots + {3^{11}}$

The terms of this sequence$$3,{3^2},{3^3}, \ldots ,{3^{11}}$$forms a G.P.

Therefore,

\begin{align}{S_n}&= \frac{{3\left[ {{{\left( 3 \right)}^{11}} - 1} \right]}}{{3 - 1}}\\&= \frac{3}{2}\left( {{3^{11}} - 1} \right)\end{align}

\begin{align}\frac{a}{r} + a + ar &= \frac{{39}}{{10}}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\left( {\frac{a}{r}} \right)\left( a \right)\left( {ar} \right) &= 1\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Hence,

$\sum\limits_{k = 1}^{11} {{3^k}} = \frac{3}{2}\left( {{3^{11}} - 1} \right)$

Substituting this value in $$\left( 1 \right)$$, we obtain

$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} = 22 + \frac{3}{2}\left( {{3^{11}} - 1} \right)$

## Chapter 9 Ex.9.3 Question 12

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.

### Solution

Let $$\frac{a}{r},a,ar$$ be the first three terms of the G.P.

It is given that

\begin{align} \frac{a}{r} + a + ar&= \frac{{39}}{{10}}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\ \left( {\frac{a}{r}} \right)\left( a \right)\left( {ar} \right)&= 1\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right) \end{align}

From , we obtain

\begin{align} {{a}^{3}}&=1 \\ \Rightarrow a&=1 \end{align}

(considering real roots)

Substituting $$a=1$$ in $$\left( 1 \right)$$, we obtain$$\left( 2 \right)$$

\begin{align}&\Rightarrow \frac{1}{r} + 1 + r = \frac{{39}}{{10}}\\&\Rightarrow 1 + r + {r^2} = \frac{{39}}{{10}}r\\&\Rightarrow 10 + 10r + 10{r^2} - 39r = 0\\&\Rightarrow 10{r^2} - 29r + 10 = 0\\&\Rightarrow 10{r^2} - 25r - 4r + 10 = 0\\&\Rightarrow 5r\left( {2r - 5} \right) - 2\left( {2r - 5} \right) = 0\\&\Rightarrow \left( {5r - 2} \right)\left( {2r - 5} \right) = 0\\&\Rightarrow r = \frac{2}{5},\frac{5}{2}\end{align}

Thus, the three terms of the G.P are $$\frac{5}{2},\;1,\;\frac{2}{5}$$

## Chapter 9 Ex.9.3 Question 13

How many terms of the G.P $$3,{{3}^{2}},{{3}^{3}},\ldots$$a$$\frac{5}{2},\;1,\;\frac{2}{5}$$re needed to give the sum $$120$$?

### Solution

The given G.P is $$3,{3^2},{3^3}, \ldots$$

Let $$n$$ terms of this G.P be required to obtain a sum as $$120$$.

Here, $$a = 3$$ and $$r = 3$$

Therefore,

\begin{align}&\Rightarrow {S_n} = \frac{{3\left( {{3^n} - 1} \right)}}{{3 - 1}}\\&\Rightarrow \frac{{3\left( {{3^n} - 1} \right)}}{2} = 120\\&\Rightarrow \frac{{120 \times 2}}{3} = {3^n} - 1\\&\Rightarrow {3^n} - 1 = 80\\&\Rightarrow {3^n} = 81\\&\Rightarrow {3^n} = {3^4}\\&\Rightarrow n = 4\end{align}

Thus, four terms of the given G.P are required to obtain the sum of $$120$$.

## Chapter 9 Ex.9.3 Question 14

The sum of first three terms of a G.P is $$16$$ and sum of the next three terms is $$128$$. Determine the first term, common ratio and sum to $$n$$ terms of the G.P.

### Solution

Let the G.P be $$a,ar,a{r^2},a{r^3}, \ldots$$

According to the given question

$$a + ar + a{r^2} = 16$$ and $$a{r^3} + a{r^4} + a{r^5} = 128$$

Therefore,

\begin{align}a\left( {1 + r + {r^2}} \right)&= 16\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\a{r^3}\left( {1 + r + {r^2}} \right)&= 128\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Dividing $$\left( 2 \right)$$ by $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{{a{r^3}\left( {1 + r + {r^2}} \right)}}{{a\left( {1 + r + {r^2}} \right)}} = \frac{{128}}{{16}}\\&\Rightarrow {r^3} = 8\\&\Rightarrow r = 2\end{align}

Substituting $$r = 2$$ in $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow a\left( {1 + 2 + 4} \right) = 16\\&\Rightarrow a\left( 7 \right) = 16\\&\Rightarrow a = \frac{{16}}{7}\end{align}

Hence,

\begin{align}{S_n}&= \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\\&= \frac{{\frac{{16}}{7}\left( {{2^n} - 1} \right)}}{{2 - 1}}\\&= \frac{{16}}{7}\left( {{2^n} - 1} \right)\end{align}

Thus, the first term, $$a = \frac{{16}}{7}$$, common ratio, $$r = 2$$ and sum to $$n$$ terms, $${S_n} = \frac{{16}}{7}\left( {{2^n} - 1} \right)$$.

## Chapter 9 Ex.9.3 Question 15

Given a G.P with $$a = 729$$ and $${7^{th}}$$ term , determine $${S_7}$$.

### Solution

It is given that $$a = 729$$ and $${a_7} = 64$$

Let $$r$$ be the common ratio of the G.P.

It is known that $${a_n} = a{r^{n - 1}}$$

Therefore,

\begin{align}&\Rightarrow {a_7} = a{r^{7 - 1}} = a{r^6}\\&\Rightarrow 64 = 729{r^6}\\&\Rightarrow {r^6} = {\left( {\frac{2}{3}} \right)^6}\\&\Rightarrow r = \frac{2}{3}\end{align}

Also,

\begin{align}{S_n}&= \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\\{S_7}&= \frac{{729\left[ {1 - {{\left( {\frac{2}{3}} \right)}^7}} \right]}}{{1 - \frac{2}{3}}}\\&= 3 \times 729\left[ {1 - {{\left( {\frac{2}{3}} \right)}^7}} \right]\\&= {\left( 3 \right)^7}\left[ {\frac{{{{\left( 3 \right)}^7} - {{\left( 2 \right)}^7}}}{{{{\left( 3 \right)}^7}}}} \right]\\&= {\left( 3 \right)^7} - {\left( 2 \right)^7}\\&= 2187 - 128\\&= 2059\end{align}

## Chapter 9 Ex.9.3 Question 16

Find a G.P for which the sum of first two terms is $$- 4$$ and the fifth term is four times the third term.

### Solution

Let $$a$$ be the first term and $$r$$ be the common ratio of the G.P.

According to the given conditions,

\begin{align}{a_5} &= 4\left( {{a_3}} \right)\\a{r^4} &= 4a{r^2}\\{r^2} &= 4\\r &= \pm 2\end{align}

Also,

${S_2} = \frac{{a\left( {1 - {r^2}} \right)}}{{1 - r}} = - 4$

Case I: for $$r = 2$$

\begin{align}&\Rightarrow \frac{{a\left[ {1 - {{\left( 2 \right)}^2}} \right]}}{{1 - 2}} = - 4\\&\Rightarrow \frac{{a\left[ {1 - 4} \right]}}{{ - 1}} = - 4\\&\Rightarrow 3a - 4\\&\Rightarrow a = - \frac{4}{3}\end{align}

Case II: for $$r = - 2$$

\begin{align}&\Rightarrow \frac{{ - a\left[ {1 - {{\left( { - 2} \right)}^2}} \right]}}{{1 - \left( { - 2} \right)}} - 4\\&\Rightarrow \frac{{a\left( {1 - 4} \right)}}{{1 + 2}} = - 4\\&\Rightarrow \frac{{ - 3a}}{3} = - 4\\&\Rightarrow a = 4\end{align}

Thus, the required G.P is $$\frac{{ - 4}}{3},\frac{{ - 8}}{3},\frac{{ - 16}}{3}, \ldots$$ or $$4, - 8,16, - 32,64 \ldots$$

## Chapter 9 Ex.9.3 Question 17

If the $${4^{th}},{10^{th}}$$and $${16^{th}}$$ terms of a G.P are $$x,\;y$$ and respectively. Prove that $$x,\;y,\;z$$ are in G.P.

### Solution

Let $$a$$ be the first term and $$r$$ be the common ratio of the G.P.

According to the given statement,

\begin{align}{a_4} = a{r^3}&= x\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{a_{10}} = a{r^9}&= y\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\{a_{16}} = a{r^{15}}&= z\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Dividing $$\left( 2 \right)$$ by $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{y}{x} = \frac{{a{r^9}}}{{a{r^3}}}\\&\Rightarrow \frac{y}{x} = {r^6}\end{align}

Dividing $$\left( 3 \right)$$ by $$\left( 2 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{z}{y} = \frac{{a{r^{15}}}}{{a{r^9}}}\\&\Rightarrow \frac{z}{y} = {r^6}\end{align}

Hence,

\begin{align}&\Rightarrow \frac{y}{x} = \frac{z}{y}\\&\Rightarrow {y^2} v= xz\\&\Rightarrow y = \sqrt {xz} \end{align}

Thus, $$x,\;y,\;z$$ are in G.P, proved.

## Chapter 9 Ex.9.3 Question 18

Find the sum to $$n$$ terms of the given sequence is $$8,88,888, \ldots$$

### Solution

The given sequence is $$8,88,888, \ldots n{\rm{ terms}}$$.

\begin{align}{S_n} &= 8 + 88 + 888 + \ldots n{\rm{ terms}}\\&= \frac{8}{9}\left[ {9 + 99 + 999 + \ldots n{\rm{ terms}}} \right]\\&= \frac{8}{9}\left[ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + \ldots n{\rm{ terms}}} \right]\\&= \frac{8}{9}\left[ {\left( {10 + {{10}^2} + {{10}^3} + \ldots n{\rm{ terms}}} \right) - \left( {1 + 1 + 1 \ldots n{\rm{ terms}}} \right)} \right]\\&= \frac{8}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} - n} \right]\\&= \frac{8}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right]\\&= \frac{{80}}{{81}}\left( {{{10}^n} - 1} \right) - \frac{8}{9}n\end{align}

## Chapter 9 Ex.9.3 Question 19

Find the sum of the products of the corresponding terms of the sequences $$2,4,8,16,32$$ and $$128,32,8,2,\frac{1}{2}$$.

### Solution

The given sequences are $$2,4,8,16,32$$ and $$128,32,8,2,\frac{1}{2}$$ .

Accordingly, the required sum,

\begin{align}S&= 2 \times 128 + 4 \times 32 + 8 \times 8 + 16 \times 2 + 32 \times \frac{1}{2}\\&= 64\left[ {4 + 2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

It can be seen $$4,2,1 \frac{1}{2},\frac{1}{{{2^2}}}$$ is a G.P.

Here, $$a=4$$ and $$r=\frac{1}{2}$$

It is known that, $${{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}$$

Therefore,

\begin{align}{S_5} &= \frac{{4\left[ {1 - {{\left( {\frac{1}{2}} \right)}^5}} \right]}}{{1 - \frac{1}{2}}}\\&= \frac{{4\left[ {1 - \frac{1}{{32}}} \right]}}{{\frac{1}{2}}}\\&= \frac{{31}}{4}\end{align}

Hence,

$\left[ {4 + 2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}}} \right] = \frac{{31}}{4}$

Putting this value in $$\left( 1 \right)$$, we obtain

\begin{align}S&= 64 \times \frac{{31}}{4}\\&= 16 \times 31\\&= 496\end{align}

Thus, the required sum is .

## Chapter 9 Ex.9.3 Question 20

Show that the products of the corresponding terms of the sequences $$a,ar,a{r^2}, \ldots a{r^{n - 1}}$$ and $$A,AR,A{R^2},A{R^{n - 1}}$$ form a G.P and find its common ratio.

### Solution

The given sequences are $$a,ar,a{{r}^{2}},\ldots a{{r}^{n-1}}$$ and $$A,AR,A{R^2},A{R^{n - 1}}$$ .

We need to prove that the sequence:$$aA,arAR,a{r^2}A{R^2}, \ldots a{r^{n - 1}}A{R^{n - 1}}$$

Let us find the ratio of the sequence

\begin{align}& \Rightarrow \frac{{{a_2}}}{{{a_1}}} = \frac{{arAR}}{{aA}} = rR\\ &\Rightarrow \frac{{{a_3}}}{{{a_2}}} = \frac{{a{r^2}A{R^2}}}{{arAR}} = rR\end{align}

Thus, the above sequence forms a G.P with common ratio $$rR$$.

## Chapter 9 Ex.9.3 Question 21

Find four numbers forming a G.P in which the third term is greater than the first term by $$9$$, the second term is greater than the $${4^{th}}$$ by $$18$$.

### Solution

Let $$a$$ be the first term and $$r$$ be the common ratio of the G.P.

Hence,

${a_1} = a,\;{a_2} = ar,\;{a_3} = a{r^2},\;{a_4} = a{r^3}$

According to the given condition,

\begin{align}&\Rightarrow {a_3} = {a_1} + 9\\&\Rightarrow a{r^2} = a + 9\\&\Rightarrow a{r^2} - a = 9\\&\Rightarrow a\left( {{r^2} - 1} \right) = 9\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\\&\Rightarrow {a_2} = {a_4} + 18\\&\Rightarrow ar = a{r^3} + 18\\&\Rightarrow a{r^3} - ar = - 18\\&\Rightarrow ar\left( {{r^2} - 1} \right) = - 18\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Dividing $$\left( 2 \right)$$ by $$\left( 1 \right)$$, we obtain

\begin{align}\frac{{ar\left( {{r^2} - 1} \right)}}{{a\left( {{r^2} - 1} \right)}} &= \frac{{ - 18}}{9}\\\Rightarrow r& = - 2\end{align}

Substituting $$r = - 2$$ in $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow a\left[ {{{\left( { - 2} \right)}^2A} - 1} \right] = 9\\&\Rightarrow a\left[ {4 - 1} \right] = 9\\&\Rightarrow 3a = 9\\&\Rightarrow a = \frac{9}{3}\\&\Rightarrow a = 3\end{align}

Thus, the first four numbers of the G.P. are $$3,3\left( { - 2} \right),3{\left( { - 2} \right)^2}$$ and $$3{\left( { - 2} \right)^3}$$

i.e., $$3, - 6,12, - 24$$.

## Chapter 9 Ex.9.3 Question 22

If $${p^{th}},{q^{th}}$$ and $${r^{th}}$$ terms of a G.P are $$a,b$$ and $$c$$ respectively. Prove that $${a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = 1$$.

### Solution

Let be the first term and $$R$$ be the common ratio of the G.P.

According to the given condition,

\begin{align}A{R^{p - 1}} &= a\\A{R^{q - 1}} &= b\\A{R^{r - 1}}&= c\end{align}

Therefore,

\begin{align}{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} &= {A^{q - r}} \times {R^{\left( {p - 1} \right)\left( {q - r} \right)}} \times {A^{r - p}} \times {R^{\left( {q - 1} \right)\left( {r - p} \right)}} \times {A^{p - q}} \times {R^{\left( {r - 1} \right)\left( {p - q} \right)}}\\&= {A^{q - r + r - p + p - q}} \times {R^{\left( {pr - pr - q + r} \right) + \left( {rq - r + p - pq} \right) + \left( {pr - p - qr + q} \right)}}\\&= {A^0} \times {R^0}\\&= 1\end{align}

Hence proved.

## Chapter 9 Ex.9.3 Question 23

If the first and $${n^{th}}$$ term of the G.P is $$a$$ and $$b$$ respectively, if $$P$$ is the product of $$n$$ terms, prove that $${P^2} = {\left( {ab} \right)^n}$$.

### Solution

It is given that the first term of the G.P is $$a$$ and the last term is $$b$$.

Let the common ratio be $$r$$

Hence, the G.P is $$a,ar,a{r^2},a{r^3}, \ldots a{r^{n - 1}}$$ and

$$b = a{r^{n - 1}}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Now, the product of $$n$$ terms

\begin{align}P& = \left( a \right) \times \left( {ar} \right) \times \left( {a{r^2}} \right) \times \ldots \times \left( {a{r^{n - 1}}} \right)\\&= \left( {a \times a \times a \ldots n{\rm{ times}}} \right)\left( {r \times {r^2} \times \ldots \times {r^{n - 1}}} \right)\\&= {a^n}{r^{1 + 2 + \ldots + \left( {n - 1} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Here, $$1,2, \ldots \left( {n - 1} \right)$$ is an A.P.

Therefore,

\begin{align}1 + 2 + \ldots + \left( {n - 1} \right) &= \frac{{n - 1}}{2}\left[ {2 + \left( {n - 1 - 1} \right) \times 1} \right]\\&= \frac{{n - 1}}{2}\left[ {2 + n - 2} \right]\\&= \frac{{n\left( {n - 1} \right)}}{2}\end{align}

Substituting this value in $$\left( 2 \right)$$, we obtain

\begin{align}P& = {a^n}{r^{\frac{{n\left( {n - 1} \right)}}{2}}}\\{P^2} &= {a^{2n}}{r^{n\left( {n - 1} \right)}}\\&= {\left[ {{a^2}{r^{\left( {n - 1} \right)}}} \right]^n}\\&= {\left[ {a \times a{r^{n - 1}}} \right]^n}\\{P^2} &= {\left( {ab} \right)^n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{using }}\left( {\rm{1}} \right)} \right]\end{align}

Thus, $${P^2} = {\left( {ab} \right)^n}$$proved.

## Chapter 9 Ex.9.3 Question 24

Show that the ratio of the sum of first $$n$$ terms of a G.P to the sum of terms from $${\left( {n + 1} \right)^{th}}$$to $${\left( {2n} \right)^{th}}$$ term is $$\frac{1}{{{r^n}}}$$.

### Solution

Let $$a$$ be the first term and $$r$$ be the common ratio of the G.P.

Sum of first $$n$$ terms, $$S = \frac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$$

Since, there are $$n$$ terms from $${\left( {n + 1} \right)^{th}}$$ to $${{\left( 2n \right)}^{th}}$$ term

Hence, sum of the $$n$$ terms from $${{\left( n+1 \right)}^{th}}$$ to $${{\left( 2n \right)}^{th}}$$ terms $${{S}_{n}}=\frac{{{a}_{n+1}}\left( 1-{{r}^{n}} \right)}{1-r}$$

It is known that $${{a}_{n}}=a{{r}^{n-1}}$$

Therefore,

\begin{align}{a_{n + 1}}& = a{r^{n + 1 - 1}}\\&= a{r^n}\end{align}

Now,

${S_n} = \frac{{a{r^n}\left( {1 - {r^n}} \right)}}{{1 - r}}$

Thus, the required ratio

\begin{align}\frac{S}{{{S_n}}} &= \frac{{\left( {\frac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}} \right)}}{{\left( {\frac{{a{r^n}\left( {1 - {r^n}} \right)}}{{1 - r}}} \right)}}\\&= \frac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}} \times \frac{{\left( {1 - r} \right)}}{{a{r^n}\left( {1 - {r^n}} \right)}}\\&= \frac{1}{{{r^n}}}\end{align}

Hence proved.

## Chapter 9 Ex.9.3 Question 25

If $$a,b,c$$ and $$d$$ are in G.P. Show that: $$\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{b^2} + {c^2} + {d^2}} \right) = {\left( {ab + bc + cd} \right)^2}$$

### Solution

If $$a,b,c$$ and $$d$$ are in G.P.

Therefore,

\begin{align}bc &= ad\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{b^2} &= ac\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\{c^2} &= bd\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

We need to prove that, $$\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{b^2} + {c^2} + {d^2}} \right) = {\left( {ab + bc + cd} \right)^2}$$

Since,

\begin{align}RHS &= {\left( {ab + bc + cd} \right)^2}\\&= {\left( {ab + ad + cd} \right)^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right)} \right]\\&= {\left[ {ab + d\left( {a + c} \right)} \right]^2}\\&= {a^2}{b^2} + 2abd\left( {a + c} \right) + {d^2}{\left( {a + c} \right)^2}\\&= {a^2}{b^2} + 2{a^2}bd + 2acbd + {d^2}\left( {{a^2} + 2ac + {c^2}} \right)\\&= {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}\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right){\rm{ and }}\left( 2 \right)} \right]\\&= {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{align}

Using (2) and (3) and rearranging the terms

\begin{align}RHS &= {a^2}\left( {{b^2} + {c^2} + {d^2}} \right) + {b^2}\left( {{b^2} + {c^2} + {d^2}} \right) + {c^2}\left( {{b^2} + {c^2} + {d^2}} \right)\\&= \left( {{a^2} + {b^2} + {c^2}} \right)\left( {{b^2} + {c^2} + {d^2}} \right)\\&= LHS\end{align}

Thus, $$\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{b^2} + {c^2} + {d^2}} \right) = {\left( {ab + bc + cd} \right)^2}$$ proved.

## Chapter 9 Ex.9.3 Question 26

Insert two numbers between $$3$$ and $$81$$ so that the resulting sequence is G.P.

### Solution

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, $$a = 3$$ and $${a_4} = 81$$

\begin{align}&\Rightarrow 3{r^3} = 81\\&\Rightarrow {r^3} = 27\\&\Rightarrow r = 3\;\;\;\;\;\;\;\;\;\;\;\left( {{\text{considering real roots only}}} \right)\end{align}

Hence,

\begin{align}{G_1} &= ar = 3 \times 3 = 9\\{G_2} &= a{r^2} = 3 \times {\left( 3 \right)^2} = 27\end{align}

Thus, the required two numbers are $$9$$ and $$27$$.

## Chapter 9 Ex.9.3 Question 27

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 $$b$$.

### Solution

It is known that G.M. of $$a$$ and $$b$$is $$\sqrt {ab}$$

By the given condition

$\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab}$

By squaring both sides, we obtain

\begin{align}&\Rightarrow \frac{{{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}^2}}}{{{{\left( {{a^n} + {b^n}} \right)}^2}}} = ab\\&\Rightarrow {a^{2n + 2}} + 2{a^{n + 1}}{b^{n + 1}} + {b^{2n + 2}} = \left( {ab} \right)\left( {{a^{2n}} + 2{a^n}{b^n} + {b^{2n}}} \right)\\&\Rightarrow {a^{2n + 2}} + 2{a^{n + 1}}{b^{n + 1}} + {b^{2n + 2}} = {a^{2n + 1}}b + 2{a^{n + 1}}{b^{n + 1}} + a{b^{2n + 1}}\\&\Rightarrow {a^{2n + 2}} + {b^{2n + 2}} = {a^{2n + 1}}b + a{b^{2n + 1}}\\&\Rightarrow {a^{2n + 2}} - {a^{2n + 1}}b = a{b^{2n + 1}} - {b^{2n + 2}}\\&\Rightarrow {a^{2n + 1}}\left( {a - b} \right) = {b^{2n + 1}}\left( {a - b} \right)\\&\Rightarrow {\left( {\frac{a}{b}} \right)^{2n + 1}} = 1 = {\left( {\frac{a}{b}} \right)^0}\\&\Rightarrow 2n + 1 = 0\\&\Rightarrow n = - \frac{1}{2}\end{align}

Thus, the value of $$n = - \frac{1}{2}$$

## Chapter 9 Ex.9.3 Question 28

The sum of two numbers is 6 times their G.M, show that numbers are in the ratio $$\left( {3 + 2\sqrt 2 } \right):\left( {3 - 2\sqrt 2 } \right)$$.

### Solution

Let the two numbers be $$a$$ and $$b$$.

Then its $$G.M. = \sqrt {ab}$$

According to the given condition,

\begin{align}&\Rightarrow a + b = 6\sqrt {ab} \;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\&\Rightarrow {\left( {a + b} \right)^2} = 36\left( {ab} \right)\end{align}

Also,

\begin{align}{\left( {a - b} \right)^2} &= {\left( {a + b} \right)^2} - 4ab\\&= 36ab - 4ab\\&= 32ab\\a - b &= \sqrt {32} \sqrt {ab} \\&= 4\sqrt 2 \sqrt {ab} \;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Adding $$\left( 1 \right)$$and $$\left( 2 \right)$$, we obtain

\begin{align}2a &= \left( {6 + 4\sqrt 2 } \right)\sqrt {ab} \\a &= \left( {3 + 2\sqrt 2 } \right)\sqrt {ab} \end{align}

Substituting the value of $$a$$ in $$\left( 1 \right)$$, we obtain

\begin{align}b &= 6\sqrt {ab} - \left( {3 + 2\sqrt 2 } \right)\sqrt {ab} \\&= \left( {3 - 2\sqrt 2 } \right)\sqrt {ab} \end{align}

Hence the ratio of the numbers is

\begin{align}\frac{a}{b} &= \frac{{\left( {3 + 2\sqrt 2 } \right)\sqrt {ab} }}{{\left( {3 - 2\sqrt 2 } \right)\sqrt {ab} }}\\&= \frac{{\left( {3 + 2\sqrt 2 } \right)}}{{\left( {3 - 2\sqrt 2 } \right)}}\end{align}

Thus, the required ratio is $$\left( {3 + 2\sqrt 2 } \right):\left( {3 - 2\sqrt 2 } \right)$$.

## Chapter 9 Ex.9.3 Question 29

If $$A$$ and $$G$$ be A.M and G.M, respectively between two positive numbers, prove that the numbers are $$A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}$$.

### Solution

It is given that $$A$$ and $$G$$are A.M and G.M, respectively between two positive numbers.

Let these two positive numbers be $$a$$ and $$b$$.

Therefore,

\begin{align}A.M&= A = \frac{{a + b}}{2}\\\Rightarrow a + b &= 2A\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

And,

\begin{align}G.M &= G = \sqrt {ab} \\\Rightarrow ab &= {G^2}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Since,

\begin{align}{\left( {a - b} \right)^2} &= {\left( {a + b} \right)^2} - 4ab\\&= 4{A^2} - 4{G^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{using }}\left( 1 \right){\rm{ and }}\left( 2 \right)} \right]\\&= 4\left( {{A^2} - {G^2}} \right)\\&= 4\left( {A + G} \right)\left( {A - G} \right)\\\left( {a - b} \right) &= 2\sqrt {\left( {A + G} \right)\left( {A - G} \right)} \;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

By adding and $$\left( 3 \right)$$, we obtain

\begin{align} & \Rightarrow 2a=2A+2\sqrt{\left( A+G \right)\left( A-G \right)} \\ & \Rightarrow a=A+\sqrt{\left( A+G \right)\left( A-G \right)} \\ \end{align}

Substituting the value of $$a$$ in $$\left( 1 \right)$$, we obtain

\begin{align}b &= 2A - \left( {A + \sqrt {\left( {A + G} \right)\left( {A - G} \right)} } \right)\\&= A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)} \end{align}

Thus, the two numbers are $$A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}$$.

## Chapter 9 Ex.9.3 Question 30

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 $${2^{nd}}$$ hour, $${4^{th}}$$ hour and $${n^{th}}$$ hour?

### Solution

It is given that the number of bacteria doubles every hour.

Hence, the number of bacteria after every hour will form a G.P with $$a = 30$$ and $$r = 2$$

Therefore, the number of bacteria at the end of $${2^{nd}}$$ hour will be

\begin{align}{a_3} &= a{r^2}\\&= 30 \times {\left( 2 \right)^2}\\&= 120\end{align}

The number of bacteria at the end of $${4^{th}}$$ hour will be

\begin{align}{a_5}&= a{r^4}\\&= 30 \times {\left( 2 \right)^4}\\&= 480\end{align}

The number of bacteria at the end of $${n^{th}}$$ hour will be

\begin{align}{a_{n + 1}} &= a{r^n}\\&= 30 \times {\left( 2 \right)^n}\end{align}

## Chapter 9 Ex.9.3 Question 31

What will $$&#8377$$ $$500$$ amount to in $$10$$ years after its deposit in a bank which pays annual interest rate of $$10\%$$ compounded annually?

### Solution

The amount deposited in the bank is $$&#8377$$ $$500$$.

At the end of first year, amount in $$&#8377$$ $$= 500\left( {1 + \frac{1}{{10}}} \right) = 500\left( {1.1} \right)$$

At the end of second year, amount in $$&#8377$$ $$= 500\left( {1.1} \right)\left( {1.1} \right)$$

At the end of third year, amount in $$&8377$$ $$= 500\left( {1.1} \right)\left( {1.1} \right)\left( {1.1} \right)$$

and so on…

Therefore, at the end of $$10$$ years, amount in ?

\begin{align}&= 500\left( {1.1} \right)\left( {1.1} \right)\left( {1.1} \right) \ldots 10{\rm{ times}}\\&= 500{\left( {1.1} \right)^{10}}\end{align}

## Chapter 9 Ex.9.3 Question 32

If A.M and G.M are roots of a quadratic equation are $$8$$ and $$5$$, respectively, then obtain the quadratic equation.

### Solution

Let the roots of the quadratic equations be $$a$$ and $$b$$.

According to the condition,

\begin{align}&A.M = \frac{{a + b}}{2} = 8\\&\Rightarrow a + b = 16\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\&G.M = \sqrt {ab} = 5\\&\Rightarrow ab = 25\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

The quadratic equation is given by,

\begin{align}&{x^2} - x\left( {{\rm{Sum of roots}}} \right) + \left( {{\rm{Product of roots}}} \right) = 0\\&{x^2} - x\left( {a + b} \right) + \left( {ab} \right) = 0\\&{x^2} - 16x + 25 = 0\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right){\rm{ and }}\left( 2 \right)} \right]\end{align}

Thus, the required quadratic equation is $${x^2} - 16x + 25 = 0$$.

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