NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.3

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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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