S Chand CLASS 10 Chapter 9 Arithmetic and Geometric Progression Exercise 9 D

  Exercise 9 D

Question 1

Ans: (i) $\Rightarrow 3,-6,12, \ldots$ to 6 terms

Here, $a=3, r=\frac{-6}{3}=-2, n=6$

$S_{6}=\frac{a\left(1-r^{n}\right)}{1-r}$  $(\because r<1)$

$=\frac{3\left[1-(-2)^{6}\right]}{1-(-2)}$

$=\frac{3(1-64)}{1+2}$

$=\frac{3(-63)}{3}$

$=-63$

(ii) $\Rightarrow-2,-6,-18, \ldots$ to 7 terms

Here a = -2 , r =$\frac{-6}{-2}=3, n=7$ 

$S_{7}=\frac{a\left(r^{n}-1\right)}{r-1}$ $(\because r>1)$

$=\frac{-2\left(3^{7}-1\right)}{3-1}$

$\left.=\frac{-2(2187-1}{2}\right)$

$=-\left(\begin{array}{lll}2 & 18 & 6\end{array}\right)$

$=-2186$

(iii) $\frac{1}{9}, \frac{1}{3}, 1, \ldots$ to 5 terms

Here, $a=\frac{1}{9}, r=\frac{1}{3} \div \frac{1}{9}=\frac{1}{3} \times \frac{9}{1}=3$,

$n=5$

$S_{5}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{\frac{1}{9}\left(3^{5}-1\right)}{3-1}$

$=\frac{1}{9 \times 2}(243-1)$

$=\frac{1}{18} \times 242$

$=\frac{121}{9}=13 \frac{4}{9}$

(iv) $\Rightarrow 2,1, \frac{1}{2}, \ldots$ to 6 terms

Here, $a=2, r=1 \div 2=\frac{1}{2}, n=6$

$S_{6}=\frac{1(1-r n)}{1-r}$  $(\because r<1)$

$=\frac{2\left[1-\left(\frac{1}{2}\right)^{6}\right]}{1 \frac{-1}{2}}$

$=\frac{2\left[1-\frac{1}{64}\right]}{1-\frac{1}{2}}$

$=\frac{2 \times 2}{1}\left[\frac{64-1}{64}\right]$

$=4 \times \frac{63}{64}=\frac{63}{16}$

$=3 \frac{15}{16}$

(v) $\Rightarrow 1, \frac{2}{3}, \frac{4}{9}, \cdots$ to 10 terms

Here, $a=1$

$r=\frac{2}{3} \div 1=\frac{2}{3}$

$n=10$

$S_{10}=\frac{a(1-r n)}{1-r}$

$=\frac{1\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}$

$=\frac{1-\left(\frac{2}{3}\right)^{10}}{\frac{1}{3}}$

$3\left[\frac{1}{3}-\left(\frac{2}{3}\right)^{10}\right]$

(vi) $0.15,0.015,0.00015, \ldots$, 20 terms

Here, $a=0.15$,

$r=\frac{0.015}{0.15}=\frac{1}{10}=0.1$,

$n=20$

$\therefore S_{20}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}$

$=\frac{0.15}{0.9}\left[1+(0.1)^{20}\right]$

$=\frac{15}{90}\left[1+(0.1)^{20}\right]$

$=\frac{1}{6}\left[1+(0.1)^{20}\right]$

Question 2

Ans : (i) $\Rightarrow 12+6+3+1.5+\ldots$ to 10 terms

Here, $a=12$,

$r=\frac{6}{12}=\frac{1}{2}$,

$n=10$

$S_{n}=\frac{a(1-r n)}{1-r}$ $(\because r<1)$

$S_{10}=\frac{12\left[1-\left(\frac{1}{2}\right)^{10}\right]}{1-\frac{1}{2}}$

$=\frac{12}{\frac{1}{2}}\left[1-\left(\frac{1}{2}\right)^{10}\right]$

$=\frac{12 \times 2}{1}\left[1-\left(\frac{1}{2}\right)^{10}\right]$

$=24\left[1-\left(\frac{1}{2}\right)^{10}\right]$

(ii) $\Rightarrow 6-3+1 \frac{1}{2}-\frac{3}{4}+\ldots$ to 15 terms

Here, $a=6,2$

$r=\frac{-3}{6}=\frac{-1}{2}$,

$n=15$

$S_{n}=\frac{a(1-r ^{n})}{1-r}$

$S_{10}=\frac{6\left[1-\left(-\frac{1}{2}\right)^{15}\right]}{1-\left(\frac{-1}{2}\right)}$

$=\frac{6\left[1-\left(\frac{-1}{2}\right)^{15}\right]}{1+\frac{1}{2}}$

$=\frac{6}{3}\left[1-\left(\frac{-1}{2}\right)^{15}\right]$

$=\frac{8 \times 2}{3}\left[1-\left(\frac{-1}{2}\right)^{15}\right]$

$=4\left[11+\left(\frac{1}{2}\right)^{15}\right]$

 

(iii) $\Rightarrow 2+6+18+54+\ldots+1012$ terms

 Here,  a=2,

$r=\frac{6}{2}=3$

n=12

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

$S_{12}=\frac{2\left(3^{12}-1\right)}{3-1}=\frac{2(3^{2}-1)}{2}$

$=3^{12}-1$

 

(iv) $\Rightarrow 6+12+24+\ldots+1536$

Here, $a=6$

$r=\frac{12}{6}=2$,

$l=1536$

$T_{n}=l=a(r^{n}-1)$

$1536=6\left(2^{n}-1\right)$

$2^{n}-1=\frac{1536}{6}=$ 

$256=2^{8}$

(TO BE ADDED)
$\therefore n-1=8 \Rightarrow n=8+1=90$

Now ,  $s_{9}=\frac{a(r n-1)}{r-1}$ $(\because r>1)$

$=\frac{6\left(2^{9}-1\right)}{2-1}$

$=\frac{6(512-1)}{1}$

$=6 \times 511$

$=3066$

Question 3

Ans : (i)  $12+6+3+1 \frac{1}{2}+\ldots . n$ terms

Here, $a=12$

$r=\frac{1}{2}$

$S_{n}=\frac{a(1-2 n)}{1-r} \quad(\because r<1)$

$=\frac{12\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}$

$=\frac{12}{\frac{1}{2}}\left[1-\left(\frac{1}{2}\right)^{n}\right]$

$=\frac{12 \times 2}{1}\left[1-\left(\frac{1}{2}\right)^{n}\right]$

$=24\left[1-\left(\frac{1}{2}\right)^{n}\right]$

(ii)  $20-10+5-2 \frac{1}{2}+\ldots n$ terms.

$\therefore a=20$

$r=\frac{-10}{20}$

$\begin{aligned} &=\frac{-1}{2} \\ S_{h} &=\frac{a(1-r^{n})}{1-r} \end{aligned}$

$=\frac{20\left[1-\left(-\frac{1}{2}\right)^{n}\right]}{1+\frac{1}{2}}$

$=\frac{20}{\frac{3}{2}}\left[1-\left(\frac{-1}{2}\right)^{n}\right] .$

$=\frac{20 \times 2}{3}\left[1-\left(\frac{-1}{2}\right)^{n}\right] .$

$=\frac{40}{3}\left[1-\left(-\frac{1}{2}\right)^{n}\right]$

(iii) $9-3,+1-\frac{1}{3}+\ldots+$ terms.

$\therefore \quad a=9$,

$r=-\frac{3}{9}$

$=-\frac{1}{3}$

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

$=\frac{9\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}$

$=\frac{9\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{\frac{4}{3}}$

$=\frac{9 \times 3}{4}\left[1-\left(\frac{-1}{3}\right)^{n}\right]$

$=\frac{27}{4}\left[1-\left(-\frac{1}{3}\right)^{n}\right]$

(iv) $\sqrt{3}+3+3 \sqrt{3}+9+\ldots \ldots$ n terms

$\begin{aligned} \therefore & a=\sqrt{3}, \\ & r=\frac{3}{\sqrt{3}} \\ &=\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \end{aligned}$

$=\frac{3 \sqrt{3}}{(\sqrt{3})^{2}}$

$=\frac{3 \sqrt{3}}{3}$

$=\sqrt{3}$

$\begin{aligned} \therefore S_{n} &=\frac{a\left(r^{n}-1\right)}{r-1} \\ &=\frac{\sqrt{3}\left[(\sqrt{3})^{n}-1\right]}{\sqrt{3}-1} \end{aligned}$

$\begin{aligned} &=\frac{\sqrt{3}}{\sqrt{3}-1}\left[(\sqrt{3})^{h}-1\right] . \\ &=\frac{\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ &=\left[(\sqrt{3})^{n}-1\right] \\ &=\frac{3+\sqrt{3}}{3-1}\left[(\sqrt{3})^{h}-1\right] \\ &=\frac{3+\sqrt{3}}{2}\left[\sqrt{3}^{n}-1\right] . \end{aligned}$

(v) $\Rightarrow 0.9+0.09+0.009+0.0009+a . . n$ terms

Here, $a=0.9$,

$\begin{aligned} r&=\frac{0.09}{0.9} \\ &=\frac{1}{10} \\ &=0.1 \end{aligned}$

$S_{n}=\frac{a(1-r ^{n})}{1-r}$

$=\frac{0.9[1-(0.1)^{ n}]}{1-0.1}$

$=\frac{0.9}{0.9\left[1-(0.1)^{n}\right]}$

$=1-(0.1)^{ n}$

Question 4

Ans: $\Rightarrow S_{n}=\frac{a(r n-1)}{r-1}$

(i) $\Rightarrow a_{n}=1000$,

$\gamma=10$ and $n=7$,

$a_{1}$ and $s_{n}$

$a_{n}=a r^{n-1} \Rightarrow a \times 10^{7-1}$

$=1000$

$\Rightarrow a \times 10^{6}=1000$

$\Rightarrow a=\frac{1000}{10^{6}}$

$=\frac{10^{3}}{10^{6}}$

$\Rightarrow a=\frac{1}{10^{6-3}}$

$=\frac{1}{10^{3}}$

$=\frac{1}{1000}=0.001$

and $s_{n}=\frac{a(rn-1)}{r-1}=\frac{0.001\left(10^{7}-1\right)}{10^{-1}}$

$=\frac{1}{1000 \times 9}(10000000-1)=\frac{1}{9000} \times 999999$

$=111.111$

(ii) $\Rightarrow \dot{a}_{1}=5$,

$a_{n}=320$

$r=2, n$ and $s_{n}$

$\begin{aligned}&a_{n}=\operatorname{ar}^{n-1} \Rightarrow 320=5 \times 2^{n}-1 \\&2^{n-1}=\frac{320}{5}=64=26\end{aligned}$

So , n - 1 = 6

n= 6+1= 7

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

$=\frac{5\left(\frac{27-1}{2-1}\right)}{1}=5 \times 127=635$

$=\frac{5(128-1)}{1}$

$=5 \times 127=635$

(iii) n = 9 

r = 2 ,  $\mathcal{S}_{n}$ = 1022 , a & $a_{1}$

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

$\Rightarrow 1022=\frac{a(29-1)}{2-1}=\frac{a(512-1)}{1}$

$a \times 511=1022 \Rightarrow a=\frac{1022}{511}=2$

$a_{9}=a r^{n-1}=2(2)^{9-1}=2 \times 28$

$=2 \times 256=512$

Question 5

Ans:  $\Rightarrow\left\{a_{n}\right\}$ is in G.S.

$a_{1}=4, r=5, a_{6}, 5_{6}$

$a_{6}=arn^{-1}$

$=4 \times(5)^{6-1}$

$=4 \times 5^{5}$

$=4 \times 5 \times 5 \times 5 \times 5 \times 5=12500$

$S_{6}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{4\left[(5)^{6}-1\right]}{5-1}$

$=\frac{4\left(5^{6}-1\right)}{4}$

$=\frac{4(56-1)}{4}$

$=56-1=15625-1$

$=15624$

Question 6

Ans: G.P is 

3, $\frac{3}{2}, \frac{3}{4}$ .....sum = $\frac{3069}{512}$

Here, $a=3, r=$ $\frac{3}{2}$

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

$\frac{3069}{512}=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}$

$=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\frac{1}{2}}=3 \times 2\left[1-\left(\frac{1}{2}\right)^{n}\right]$

$1=\left(\frac{1}{2}\right)^{n}=\frac{3069}{512 \times 6}=\frac{1023}{1024}$

$1-\frac{1023}{1024}=\left(\frac{1}{2}\right)^{n}$

$\Rightarrow \frac{1024-1023}{1024}=\left(\frac{1}{2}\right)^{n}$

$\Rightarrow\left(\frac{1}{2}\right)^{n}=\frac{1}{1024}=\frac{1}{(2)^{10}}$

$=\left(\frac{1}{2}\right)^{10}$

$\therefore n=10$

So $\frac{3069}{512}$ is the 10th term

Question 7

Ans: Sum of some terms = 315 

Ratio in first term  $\left(a_{1}\right)$ and common ratio (r)

= 5:2

Let number of terms be n

and first term $\left(a_{1}\right)$ be 5 and r = 2

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

$315=\frac{5(2 n-1)}{2-1} \Rightarrow \frac{315}{5}=\frac{2 n-1}{1}$

$63=2^{n}-1 \Rightarrow 2 n=63+1=64=26$

$\therefore n=6$

and $T_{6} \ldots$ or $a_{6}$

$=a=arn-1=5 (2) 6-1$

$=5 \times 2^{5}$

$=5 \times 32$

$=160$

 

Question 8

Ans: In a G.P 

a= 729 

$T_{7}=64, S_{7}$

$T_{7}=ar^{n-1}$

$64=729.r^{7-1}$

$64=729.r^{6}$

$\frac{64}{729}=r^{6}$

$\frac{2^{6}}{3^{0}}=r^{6}$

$\left(\frac{2}{3}\right)^{6}=r^{6}$

On comparing 

$r=\frac{2}{3}$

Then , 

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

$=\frac{729\left[1-\left(\frac{2}{3}\right)^{7}\right]}{1-\frac{2}{3}}$

$=\frac{729\left[1-\frac{128}{2187}\right]}{\frac{3-2}{3}}$

$=\frac{729\left[\frac{2187-128}{2187}\right]}{\frac{1}{3}}$

$=729 \times 3\left[\frac{2059}{2187}\right]$

= 2059

Question 9

Ans: Given, 

G.P is 

2+ 6 + 18 + ......+ 4373

a = 2 

$r=\frac{6}{2}=3$

and $l=4374$.

Then, $a_{n}=l=a r^{n-1}$.

$4374=2 \times 3^{n-1}$

$\frac{4374}{2}=3^{n-1}$

$2187=3^{n-1}$

$3^{7}=3^{n-1}$

On comparing ,

n -1 = 7 

n = 7 +1

n= 8

$\begin{aligned} \therefore S_{8} &=\frac{a\left(r^{n}-1\right)}{r-1} \\ &=\frac{2\left(3^{8}-1\right)}{3-1} \end{aligned}$

$=\frac{2(6561-1)}{2}$

= 6560

Question 10

Ans:  Given, 

The G.P is 

$\sqrt{3}, 3,3 \sqrt{3}, \ldots$ and $S_{n}=39+13 \sqrt{3}$.

$\therefore a=\sqrt{3}, r=\sqrt{3}$

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

$39+13 \sqrt{3}=\frac{\sqrt{3}\left[(\sqrt{3})^{n}-1\right]}{\sqrt{3}-1}$

$(\sqrt{3}-1)(39+13 \sqrt{3})=\sqrt{3}\left[(\sqrt{3})^{n}-1\right]$

$\frac{(\sqrt{3}-1)(39+13 \sqrt{3})}{\sqrt{3}}=(\sqrt{3})^{n}-1$

$-(13 \sqrt{3}+13)(\sqrt{3}-1)$

$=13(\sqrt{3}+1)(\sqrt{3}-1)$

$=13(3-1) .$

$=13 \times 2$

$=26$

So,  $(\sqrt{3})^{n}=26+1$

$(\sqrt{3})^{n}=27$

$(\sqrt{3})^{n}=3^{3}$

$(\sqrt{3})^{n}=(\sqrt{3})^{6}$

On comparing , 

n = 6 

Hence , Number of terms = 6