Exercise 8.2 (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths

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Chapter 8 - Sequences & Series | NCERT Solutions for Class 11 Maths

Ex 8.2 Question 1.

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

Answer.

Here $a=\frac{5}{2}$ and $\gamma=\frac{5}{4} \div \frac{5}{2}=\frac{1}{2}$
$
\begin{aligned}
& \therefore a_n=a r^{n-1} \\
& \Rightarrow a_{20}=\frac{5}{2} \times\left(\frac{1}{2}\right)^{20-1} \\
& \Rightarrow a_{20}=\frac{5}{2} \times\left(\frac{1}{2}\right)^{19}=\frac{5}{2^{20}}
\end{aligned}
$

And $a_n=\frac{5}{2} \times\left(\frac{1}{2}\right)^{n-1}=\frac{5}{2 \times 2^{n-1}}=\frac{5}{2^n}$
Ex 8.2 Question 2.

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

Answer.

Let $a$ be the first term of given G.P. Here $r=2$ and $a_8=192$
$
\begin{aligned}
& \therefore a_n=a r^{n-1} \\
& \Rightarrow a_8=a \times(2)^{8-1}=192
\end{aligned}
$

$\begin{aligned}
& \Rightarrow a \times(2)^7=192 \\
& \Rightarrow a \times 128=192
\end{aligned}$

$\begin{aligned}
& \Rightarrow a=\frac{192}{128}=\frac{3}{2} \\
& \therefore a_{12}=a r^{12-1} \\
& \Rightarrow a_{12}=\frac{3}{2} \times 2^{11}=3 \times 2^{10} \\
& =3 \times 1024=3072
\end{aligned}$

Ex 8.2 Question 3.

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

Answer.

Let $a$ be the first term and $\gamma$ be the common ratio of given G.P.
$
\begin{aligned}
& \therefore a_5=p \Rightarrow a r^4=p \\
& a_{\mathbb{B}}=q \Rightarrow a r^7=q \cdots \\
& a_{11}=s \Rightarrow a r^{10}=s \cdots
\end{aligned}
$

Squaring both sides of eq. (ii), we get $q^2=\left(a r^7\right)^2$
$
\begin{aligned}
& \Rightarrow q^2=a^2 r^{14} \\
& \Rightarrow q^2=\left(a r^4\right)\left(a r^{10}\right) \\
& \Rightarrow q^2=p s \quad \text { [From eq. (i) and (iii)] }
\end{aligned}
$
Ex 8.2 Question 4.

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

Answer.

Let $a$ be the first term and $\gamma$ be the common ratio of given G.P.
Here $a=-3$ and $a_4=\left(a_2\right)^2$3

$\text { Now, } a_4=\left(a_2\right)^2$

$
\begin{aligned}
& \Rightarrow a r^3=(a r)^2 \\
& \Rightarrow a r^3=a^2 r^2 \\
& \Rightarrow r=a \\
& \Rightarrow r=-3[\because a=-3] \\
& \therefore a_7=a r^{7-1}=(-3) \times(-3)^6 \\
& =-3 \times 729=-2187
\end{aligned}
$
Ex 8.2 Question 5.

Which term of the following sequences:
(a) $2,2 \sqrt{2}, 4, \ldots \ldots$ is 128 ?
(b) $\sqrt{3}, 3,3 \sqrt{3}$ : is 729 ?
(c) $\frac{1}{3}=\frac{1}{9}=\frac{1}{27}, \ldots \ldots$ is $\frac{1}{19683}$ ?

Answer.

(a) Here $a=2, r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$ and $a_n=128$
$
\therefore a_n=a r^{n-1}
$

$\begin{aligned}
& \Rightarrow 128=2 \times(\sqrt{2})^{n-1} \\
& \Rightarrow 64=(\sqrt{2})^{n-1} \\
& \Rightarrow(\sqrt{2})^{12}=(\sqrt{2})^{n-1} \\
& \Rightarrow n-1=12 \\
& \Rightarrow n=13
\end{aligned}$

Therefore, $13^{\text {th }}$ term of the given G.P. is 128 .
(b) Here $a=\sqrt{3}, r=\frac{3}{\sqrt{3}}=\sqrt{3}$ and $a_n=729$
$
\begin{aligned}
& \therefore a_n=a r^{n-1} \\
& \Rightarrow 729=\sqrt{3} \times(\sqrt{3})^{n-1} \\
& \Rightarrow(\sqrt{3})^{12}=(\sqrt{3})^n \\
& \Rightarrow n=12
\end{aligned}
$

Therefore, $12^{\text {th }}$ term of the given G.P. is 729 .
(c) Here $a=\frac{1}{3}, r=\frac{1}{9} \div \frac{1}{3}=\frac{1}{3}$ and $a_n=\frac{1}{19683}$
$
\begin{aligned}
& \therefore a_n=a r^{n-1} \\
& \Rightarrow \frac{1}{19683}=\frac{1}{3} \times\left(\frac{1}{3}\right)^{n-1} \\
& \Rightarrow\left(\frac{1}{3}\right)^9=\left(\frac{1}{3}\right)^n
\end{aligned}
$

$
\Rightarrow n=9
$

Therefore, $9^{\text {th }}$ term of the given G.P. is $\frac{1}{19683}$.
Ex 8.2 Question 6.

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

Answer.

Given: $\frac{-2}{7}, x, \frac{-7}{2}$ are in G.P.

$
\begin{aligned}
& \therefore \frac{x}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{x} \\
& \Rightarrow x^2=\frac{-2}{7} \times \frac{-7}{2} \\
& \Rightarrow x^2=1 \\
& \Rightarrow x= \pm 1
\end{aligned}
$

Therefore for $x= \pm 1$ the given numbers are in G.P

Find the sum to indicated number of terms in each of the geometric progression in Exercises 7 to 10 :
Ex 8.2 Question 7.

$0.15,0.015,0.0015$, $\qquad$ .20 terms

Answer.

Here, $a=0.15$ and $\gamma=\frac{0.015}{0.15}=\frac{1}{10}$
$
\therefore \mathrm{S}_n=\frac{a\left(1-\gamma^n\right)}{1-\gamma} \text { when } \gamma<1
$

$
\begin{aligned}
& \Rightarrow \mathrm{S}_{20}=\frac{0.15\left[1-\left(\frac{1}{10}\right)^{20}\right]}{1-\frac{1}{10}} \\
& \Rightarrow S_{20}=\frac{15}{100} \times \frac{10}{9}\left[1-(0.1)^{20}\right] \\
& \Rightarrow S_{20}=\frac{1}{6}\left[1-(0.1)^{20}\right]
\end{aligned}
$
Ex 8.2 Question 8.

$\sqrt{7}=\sqrt{21}, 3 \sqrt{7}$, n terms

Answer.

Here, $a=\sqrt{7}$ and $r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1} \text { when } r>1 \\
& \Rightarrow \mathrm{S}_n=\frac{\sqrt{7}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1} \\
& \Rightarrow \mathrm{S}_n=\frac{\sqrt{7}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\left[(3)^{\frac{n}{2}}-1\right] \\
& \Rightarrow \mathrm{S}_n=\frac{\sqrt{7}(\sqrt{3}+1)}{2}\left[(3)^{\frac{n}{2}}-1\right]
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \mathrm{S}_n=\frac{1\left[1-(-\alpha)^n\right]}{1-(-\alpha)} \\
& \Rightarrow \mathrm{S}_n=\frac{1}{1+\alpha}\left[1-(-\alpha)^n\right]
\end{aligned}
$
Ex 8.2 Question 10.

$x^3, x^5 x^7$ $\qquad$ $n$ terms (if $x \neq \pm 1$ )

Answer.

Here, $a=x^3$ and $y^2=\frac{x^5}{x^3}=x^2$

$\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(1-r^n\right)}{1-r} \text { when } r<1 \\
& \Rightarrow \mathrm{S}_n=\frac{x^3\left[1-\left(x^2\right)^n\right]}{1-x^2} \\
& \Rightarrow \mathrm{S}_n=\frac{x^3}{1-x^2}\left[1-x^{2 n}\right]
\end{aligned}$

Ex 8.2 Question 11.

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

Answer.

Given: $\sum_{k=1}^{11}\left(2+3^k\right)$
$
=\left(2+3^1\right)+\left(2+3^2\right)+\left(2+3^3\right)+\ldots \ldots+\left(2+3^{11}\right)
$

$\begin{aligned}
& =\left(2+3^1\right)+\left(2+3^2\right)+\left(2+3^3\right)+\ldots \ldots .+\left(2+3^{11}\right) \\
& =(2+2+2+\ldots \ldots 11 \text { times })+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right)
\end{aligned}$

$
=22+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right)
$

Here $3,3^2, 3^3, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots, 3^{11}$ is in G.P.

$
\begin{aligned}
& \therefore a=3 \text { and } \gamma=\frac{3^2}{3}=3 \\
& \therefore S_n=\frac{3\left(3^{11}-1\right)}{3-1}=\frac{3}{2}\left(3^{11}-1\right)
\end{aligned}
$

Putting the value of $\mathrm{S}_n$ in eq. (i), we get $\sum_{k=1}^{11}\left(2+3^k\right)=22+\frac{3}{2}\left(3^{11}-1\right)$

$\text { 12. The sum of first three terms of a G.P. is } \frac{39}{10} \text { and their product is } 1 \text {. Find the common }$

ratio and the terms.

Answer.

Let $\frac{a}{\gamma}, a, \gamma$ be first three terms of the given G.P.
According to question, $\frac{a}{\gamma}+a+a r=\frac{39}{10}$
$
\begin{aligned}
& \text { And } \frac{a}{r} \times a \times r=1 \\
& \Rightarrow a^3=1 \\
& \Rightarrow a=1
\end{aligned}
$

Putting value of $a$ in eq. (i), $\frac{1}{r}+1+r=\frac{39}{10}$
$
\begin{aligned}
& \Rightarrow 10+10 r+10 r^2=39 r \\
& \Rightarrow 10 r^2-29 r+10=0 \\
& \Rightarrow r=\frac{-(-29) \pm \sqrt{(-29)^2-4 \times 10 \times 10}}{2 \times 10} \\
& \Rightarrow r=\frac{29 \pm \sqrt{841-400}}{20}
\end{aligned}
$

$
\Rightarrow \gamma=\frac{29 \pm 21}{20}
$

Taking $r=\frac{29+21}{20}=\frac{50}{20}=\frac{5}{2}$ and Taking $r=\frac{29-21}{20}=\frac{8}{20}=\frac{2}{5}$
When $\gamma=\frac{5}{2}$, then first three terms are $\frac{1}{5 / 2}, 1,1 \times \frac{5}{2}$

$
\Rightarrow \frac{2}{5}, 1, \frac{5}{2}
$

When $r=\frac{2}{5}$, then first three terms are $\frac{1}{2 / 5}=1,1 \times \frac{2}{5}$
$
\Rightarrow \frac{5}{2}, 1, \frac{2}{5}
$
Ex 8.2 Question 13.

How many terms of G.P. $3,3^2, 3^3$ are needed to give the sum 120 ?

Answer.

Here, $\therefore a=3$ and $\gamma=\frac{3^2}{3}=3$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1} \text { when } r>1 \\
& \Rightarrow 120=\frac{3\left(3^n-1\right)}{3-1} \\
& \Rightarrow 120=\frac{3}{2}\left(3^n-1\right) \\
& \Rightarrow 120 \times \frac{2}{3}=3^n-1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 3^n=81 \\
& \Rightarrow 3^n=(3)^4 \\
& \Rightarrow n=4
\end{aligned}
$

Therefore, the sum of 4 terms of the given G.P. is 120 .
Ex 8.2 Question 14.

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

NA. Let $a$ be the first term and $r$ be the common ratio of given G.P.
$
\begin{aligned}
& \therefore a+a r+a r^2=16 \\
& \Rightarrow a\left(1+r+r^2\right)=16
\end{aligned}
$

And $a r^3+a r^4+a r^5=128$
$
\Rightarrow a r^3\left(1+\mu+\mu^2\right)=128
$

Putting the value from eq. (i) into eq. (ii), we get
$
\begin{aligned}
& 16 r^3=128 \\
& \Rightarrow r^3=8 \\
& \Rightarrow r^3=2
\end{aligned}
$

Putting value of $r$ in eq. (i), we get $a\left(1+2+2^2\right)=16$
$
\begin{aligned}
& \Rightarrow a=\frac{16}{7} \\
& \therefore \mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1} \text { when } r>1
\end{aligned}
$

$
\Rightarrow \mathrm{S}_n=\frac{\frac{16}{7}\left(2^n-1\right)}{2-1}=\frac{16}{7}\left(2^n-1\right)
$
Ex 8.2 Question 15.

Given a G.P. with $a=729$ and $7^{\text {th }}$ term 64 , determine $S_7$.

Answer.

Given: $a=729$ and $a_7=64$
$
\begin{aligned}
& \Rightarrow a r^6=64 \\
& \Rightarrow 729 r^6=64
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow r^6=\frac{64}{729}=\left(\frac{2}{3}\right)^6 \\
& \Rightarrow r=\frac{2}{3} \\
& \therefore \mathrm{S}_n=\frac{a\left(1-r^n\right)}{1-r} \text { when } r<1 \\
& \Rightarrow \mathrm{S}_7=\frac{729\left[1-\left(\frac{2}{3}\right)^7\right]}{1-\frac{2}{3}}=\frac{729\left[1-\frac{128}{2187}\right]}{\frac{3-2}{3}} \\
& \Rightarrow \mathrm{S}_7=729 \times 3\left(\frac{2187-128}{2187}\right) \\
& \Rightarrow \mathrm{S}_7=\frac{729 \times 3 \times 2059}{2187}=2059
\end{aligned}
$
Ex 8.2 Question 16.

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

Answer.

Let $a$ be the first term and $r$ be the common ratio of given G.P.
Given: $a+a^r=-4$

$
\Rightarrow a(1+r)=-4
$

And $a_5=4 a_3$
$
\begin{aligned}
& \Rightarrow a r^4=4 a r^2 \\
& \Rightarrow r^2=4 \\
& \Rightarrow r= \pm 2
\end{aligned}
$

Putting $r=2$ in eq. (i), we get $a(1+2)=-4$

$
\Rightarrow a=\frac{-4}{3}
$

Therefore, required G.P. is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots$
Putting $r=-2$ in eq. (i), we get $a(1-2)=-4$
$
\Rightarrow a=4
$

Therefore, required G.P. is $4,-8,16,-32$.

Ex 8.2 Question 17.

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

Answer.

Let $a$ be the first term and $\gamma$ be the common ratio of given G.P.
$
\begin{aligned}
& \therefore a_4=x \\
& \Rightarrow a r^3=x \\
& a_{10}=y \\
& \Rightarrow a r^9=y
\end{aligned}
$
$
\begin{aligned}
& a_{16}=z \\
& \Rightarrow a r^{15}=z
\end{aligned}
$

From eq. (ii), $a r^9=y$
$
\begin{aligned}
& \Rightarrow\left(a r^9\right)^2=y^2 \\
& \Rightarrow y^2=\left(a r^3\right)\left(a p^{15}\right) \\
& \Rightarrow y^2=x z \text { [From eq. (i) and (iii)] } \\
& \therefore x y=z \text { are in G.P. }
\end{aligned}
$

Ex 8.2 Question 18.

Find the sum to $n$ terms of the sequences $8,88,888,8888, \ldots .$.

Answer.

Here $\mathrm{S}_n=8+88+888+8888+$ $\qquad$ up to $n$ terms
$
\begin{aligned}
& \Rightarrow \mathrm{S}_n=8(1+11+111+1111+\ldots . . \text { up to } n \text { terms }) \\
& \Rightarrow \mathrm{S}_n=\frac{8}{9}(9+99+999+9999+\ldots \text { up to } n \text { terms }) \\
& \Rightarrow \mathrm{S}_n=\frac{8}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\ldots . . \text { up to } n \text { terms }\right] \\
& \Rightarrow \mathrm{S}_n=\frac{8}{9}\left[\left(10+10^2+10^3+\ldots \text { up to } n \text { terms }\right)-(1+1+1+\ldots \text { up to } n \text { terms })\right] \\
& \Rightarrow \mathrm{S}_n=\frac{8}{9}\left[\frac{10 \times\left(10^n-1\right)}{10-1}-n\right] \\
& =\frac{8}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right] \\
& =\frac{80}{81}\left(10^n-1\right)-\frac{8}{9} n
\end{aligned}
$

Ex 8.2 Question 19.

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

Answer.

Multiplying the corresponding terms of the given sequences 2, 4, 8,16, 32 and 128, 32, 8, $2, \frac{1}{2}$
$(2 \times 128),(4 \times 32),(8 \times 8),(16 \times 2),\left(32 \times \frac{1}{2}\right)$
$\Rightarrow 256,128,64,32,16$ are in G.P.

Here $a=256, r=\frac{128}{256}=\frac{1}{2}$ and $n=5$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(1-r^n\right)}{1-r} \text { when } r<1 \\
& \Rightarrow \mathrm{S}_5=\frac{256\left[1-\left(\frac{1}{2}\right)^5\right]}{1-\frac{1}{2}}=256 \times 2\left(1-\frac{1}{32}\right) \\
& \Rightarrow \mathrm{S}_5=256 \times 2 \times \frac{31}{32}=496
\end{aligned}
$

Ex 8.2 Question 20.

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

Answer.

Multiplying the corresponding terms of the given sequences, we have
$
(a \times \mathrm{A}),(a r \times \mathrm{AR}),\left(a r^2 \times \mathrm{AR}^2\right)=\ldots \ldots \ldots \ldots \ldots,\left(a r^{n-1} \times \mathrm{AR}^{n-1}\right)
$

$\Rightarrow(a \mathrm{~A}),(a \mathrm{~A} r \mathrm{R}),\left(a \mathrm{~A} r^2 \mathrm{R}^2\right), \ldots \ldots \ldots \ldots \ldots,\left(a \mathrm{~A} r^{n-1} \mathrm{R}^{n-1}\right)$ are in G.P.
Now $\frac{a_2}{a_1}=\frac{a A r R}{a A}=r R a n d \frac{a_3}{a_2}=\frac{a A r^2 R^2}{a A r R}=r R$
Since the ratio of the two succeeding terms are same, the resulting sequence is also in G.P
$
\text { and common ratio }=\frac{a \mathrm{~A} r \mathrm{R}}{a \mathrm{~A}}=r \mathrm{R}
$
Ex 8.2 Question 21.

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than by $4^{\text {th }}$ by 18 .

Answer.

Let the four numbers in G.P. be $a, a r, a r^2: a r^3$

$
\therefore a r^2=a+9 \text { and } a r=a r^3+18
$

Now, $a r^2-a=9$
$
\Rightarrow a\left(r^2-1\right)=9
$

And $a r-a r^3=18$
$
\begin{aligned}
& \Rightarrow \operatorname{ar}\left(1-\mu^2\right)=18 \\
& \Rightarrow-a r\left(r^2-1\right)=18
\end{aligned}
$

Dividing eq. (ii) by eq. (i), we have
$
\begin{aligned}
& \frac{-a r\left(r^2-1\right)}{a\left(r^2-1\right)}=\frac{18}{9} \\
& \Rightarrow r=-2
\end{aligned}
$

Putting value of $\gamma$ in eq. (i), we get
$
\begin{aligned}
& a(4-1)=9 \\
& \Rightarrow a=3 \\
& \therefore a r=3 \times(2)=-6
\end{aligned}
$

$
\begin{aligned}
& a r^2=3 \times(-2)^2=12 \\
& a r^3=3 \times(-2)^3=-24
\end{aligned}
$

Therefore, the required numbers are $3,-6,12,-24$.
Ex 8.2 Question 22.

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

Answer.

Let $\mathrm{A}$ be the first term and $\mathrm{R}$ be the common ratio of given G.P.
$
\begin{aligned}
& \therefore a_p=a \\
& \Rightarrow \mathrm{AR}^{y-1}=a \\
& a_q=b \\
& \Rightarrow \mathrm{AR}^{q-1}=b \\
& a_r=c \\
& \Rightarrow \mathrm{AR}^{r-1}=c
\end{aligned}
$

Ex 8.2 Question 23.

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

Answer.

Let $r$ be the common ratio of the given G.P
Here, first term of G.P. is $a$
and $a_n=b$
$
\Rightarrow a r^{n-1}=b
$
$\qquad$ $a r^{n-1}$
$\Rightarrow \mathrm{P}=a^n \cdot r^{1+2+3+\ldots+n-1}$
$\Rightarrow n=a^n r \frac{n(n-1)}{2}$

$\Rightarrow p^2=a^{2 n} r^{n(n-1)}=\left[a a r^{n-1}\right]^n$ [Squaring both sides]
$\Rightarrow P^2=(a b)^n$ [From eq. (i)]
Hence proved

Ex 8.2 Question 24.

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

Answer.

Let $a$ be the first term and $\gamma$ be the common ratio of given G.P.

Ex 8.2 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)=(a b+b c+c d)^2 \text {. }
$

Answer.

Given a,b,c,d are in G.P
Let $r$ be the common ratio of given G.P.
Then $b=a r^* c=a r^2$ and $d=a r^3$
Now, L.H.S. $=\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)$
$
\begin{aligned}
& =\left(a^2+a^2 r^2+a^2 r^4\right)\left(a^2 r^2+a^2 r^4+a^2 r^6\right) \\
& =a^2\left(1+r^2+r^4\right) a^2 r^2\left(1+r^2+r^4\right)=a^4 \gamma^2\left(1+r^2+r^4\right)^2
\end{aligned}
$

$
\begin{aligned}
& \text { R.H.S. }=(a b+b c+c d)^2 \\
& =\left(a a r+a r \cdot a r^2+a r^2 \cdot a r^3\right)^2 \\
& =\left(a^2 r+a^2 r^3+a^2 r^5\right)^2 \\
& =\left(a^2 r\right)^2\left(1+r^2+r^4\right)^2=a^4 r^2\left(1+r^2+r^4\right)^2
\end{aligned}
$

Therefore, L.H.S. = R.H.S.

Ex 8.2 Question 26.

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

Answer.

Let $G_1$ and $G_2$ be two numbers between 3 and 81 such that $3, G_1, G_2, 81$ are in G.P.
Let $r$ be the common ratio
Here $a=3$ and $a_4=81$
$
\begin{aligned}
& \Rightarrow a r^3=81 \\
& \Rightarrow 3 \times r^3=81 \\
& \Rightarrow r^3=27 \\
& \Rightarrow r=3 \\
& \therefore G_1=a r=3 \times 3=9
\end{aligned}
$

And $G_2=a r^2=3 \times(3)^2=27$
Therefore, the required numbers are 9 and 27 .
Ex 8.2 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 $b$

Answer.

Since, G.M. between two numbers $a$ and $b$ is $\sqrt{a b}$.
According to question, $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{a b}$
$
\begin{aligned}
& \Rightarrow \frac{a^{n+1}+b^{n+1}}{a^n+b^n}=a^{\frac{1}{2}} b^{\frac{1}{2}} \\
& \Rightarrow a^{n+1}+b^{n+1}=\left(a^n+b^n\right) a^{\frac{1}{2}} b^{\frac{1}{2}} \\
& \Rightarrow a^{n+1}+b^{n+1}=a^{n+\frac{1}{2}} b^{\frac{1}{2}}+a^{\frac{1}{2}} b^{n+\frac{1}{2}} \\
& \Rightarrow a^{n+1}-a^{n+\frac{1}{2}} b^{\frac{1}{2}}=a^{\frac{1}{2}} b^{n+\frac{1}{2}}-b^{n+1} \\
& \Rightarrow a^{n+\frac{1}{2}}\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)=b^{n+\frac{1}{2}}\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right) \\
& \Rightarrow a^{n+\frac{1}{2}}=b^{n+\frac{1}{2}} \\
& \Rightarrow \frac{a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}=1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow\left(\frac{a}{b}\right)^{n+\frac{1}{2}}=\left(\frac{a}{b}\right)^0 \\
& \Rightarrow n+\frac{1}{2}=0 \\
& \Rightarrow n=-\frac{1}{2}
\end{aligned}
$
Ex 8.2 Question 28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.

Answer.

Let the numbers be $a$ and $b$
Given: $a+b=6 \sqrt{a b} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{3}{1}$
Applying componendo and dividendo, we get
$
\begin{aligned}
& \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{3+1}{3-1} \\
& \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{4}{2} \\
& \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}
\end{aligned}
$

Again applying componendo and dividendo, we get
$
\begin{aligned}
& \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \\
& \Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}
\end{aligned}
$

Squaring both sides, $\frac{a}{b}=\frac{2+1+2 \sqrt{2}}{2+1-2 \sqrt{2}}$
$
\Rightarrow \frac{a}{b}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}
$

Therefore, the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.
Ex 8.2 Question 29.

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

Answer.

Let the two positive numbers be $\mathrm{a}$ and $\mathrm{b}$ Therefore $\mathrm{A}=\frac{a+b}{2}$ and $\mathrm{G}=\sqrt{a b}$
$
\begin{aligned}
& \text { Now, } \mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}=\mathrm{A} \pm \sqrt{\mathrm{A}^2-\mathrm{G}^2} \\
& =\frac{a+b}{2} \pm \sqrt{\left(\frac{a+b}{2}\right)^2-(\sqrt{a b})^2} \\
& =\frac{a+b}{2} \pm \sqrt{\frac{a^2+b^2+2 a b}{4}-a b} \\
& =\frac{a+b}{2} \pm \sqrt{\frac{a^2+b^2+2 a b-4 a b}{4}}
\end{aligned}
$

$
\begin{aligned}
& =\frac{a+b}{2} \pm \sqrt{\frac{(a-b)^2}{4}}=\frac{a+b}{2} \pm \frac{a-b}{2} \\
& =\frac{a+b}{2}+\frac{a-b}{2} \text { and } \frac{a+b}{2}-\frac{a-b}{2} \\
& =\frac{a+b+a-b}{2} \text { and } \frac{a+b-a+b}{2} \\
& =\frac{2 a}{2}=a \text { and } \frac{2 b}{2}=b
\end{aligned}
$
Ex 8.2 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 $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {nh }}$ hour?

Answer.

Bacteria present in the culture originally $=30$

Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.

Here $a=30$ and $\gamma=2$
$\therefore$ Bacteria at the end of $2^{\text {nd }}$ hour $=30 \times 2^{3-1}=30 \times 2^2=120$

And Bacteria at the end of $4^{\text {th }}$ hour $=30 \times 2^{5-1}=30 \times 2^4=480$
And Bacteria at the end of $n^n$ hour $=a_{n+1}=30\left(2^{(n+1)-1}\right)=30\left(2^n\right)$
Ex 8.2 Question 31.

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

Answer.

Original amount $=$ Rs. 500 , Rate of interest $=10 \%$ compounded annually
$\therefore$ Interest of one year $=\frac{500 \times 10 \times 1}{100}=$ Rs. 50
And Amount after one year $=500+50=$ Rs. 550
Here $a=500$ and $r=\frac{550}{500}=1.1$

Therefore, amount after 10 years =Amount in the $11^{\text {th }}$ year=
$
500 \times(1.1)^{11-1}=\operatorname{Rs} .500(1.1)^{10}
$

Ex 8.2 Question 32.

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

Answer.

Let $a$ and $b$ be the roots of required quadratic equation.
Then A.M. $=\frac{a+b}{2}=8$
$
\Rightarrow a+b=16
$

And G.M. $=\sqrt{a b}=5$
$
\Rightarrow a b=25
$

Now, Quadratic equation $x^2-($ Sum of roots $) x+($ Product of roots $)=0$
$
\begin{aligned}
& \Rightarrow x^2-(a+b) x+a b=0 \\
& \Rightarrow x^2-16 x+25=0
\end{aligned}
$

Therefore, required equation is $x^2-16 x+25=0$.