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

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

  Exercise 9 C

Question 1 

Ans: (i) $27,9,3,1, \ldots$

$\begin{aligned}\therefore & a=27 \\r &=\frac{9}{27}=\frac{1}{3}, \frac{3}{9} \\&=\frac{1}{3}, .....\end{aligned}$

Hence, it is a G.P. and $r=\frac{1}{3}$

(ii) $-1,2,4,8, \ldots .$

$\therefore a=-1$,

$\quad b=\frac{2}{-1}=-2, \frac{4}{2} .$

$\quad=2, \frac{8}{4}=2 .$

So, it is not a G.P 

(iii) 

$\begin{aligned} & 2, \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \ldots \\ \therefore & a=2, \\ &=\frac{1}{2} \div 2 \\ &=\frac{1}{2} \times \frac{1}{2} \\ &=\frac{1}{4} . \\ \frac{1}{8} &=\frac{1}{2} . \end{aligned}$

$=\frac{1}{4} .$

$\frac{1}{32}=\frac{1}{8} .$

$=\frac{1}{32} \times 8 .$

$=\frac{1}{4}$

∴ IT is GP and  $r=\frac{1}{4}$

$(i v)-12,-6,0,6, \ldots . .$

$\begin{aligned}&\therefore a=-13 \\&r=\frac{-6}{-12}=\frac{1}{2} \\&=\frac{0}{-6} \\&=0\end{aligned}$

Hence,  it is not G.P.

Question 2

Ans:  (i) 2,6....

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

2,6,18,54,162

(ii) $\frac{1}{16},-\frac{1}{8}, \ldots$

$\left(r=-\frac{1}{8} \div \frac{1}{10}=-\frac{1}{8} \times 16=-2 .\right)$

$\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}, 1$

(iii)0.3, 0.06......

$r=\frac{0.01}{0.3}$= $\frac{1}{5}=0-2$

$0.3,0.03,0.012,0.0024,0.00048$

Question 3

Ans: 6th term of the G.P 2, 10 , 50......

∴ a = 2 

r=  $\frac{10}{2}=5$

$\begin{aligned} So  T_{6} &=9 x^{h-1} \\ &=2 \times 5^{6-1} \\ &=2 \times 5^{5} \\ &=2 \times 3125 \\ &=6250 \end{aligned}$

(ii)  11th term of the G.P. $4,12,3, \ldots .$

$\begin{aligned} \therefore & a=4 \\ & r=\frac{12}{4}=3 . \end{aligned}$

$\begin{aligned} T_{11} &=ar^{h-1} \\ &=4 \times(3)^{11-1} \\ &=4 \times 3^{10} \\ &=4 \times 59,049 \\ &=236196 \end{aligned}$

Question 4

Ans:

 (i) $\begin{aligned} T_{n} &=4 \cdot 3^{h-1} \\ \therefore T_{1} &=4 \cdot 3^{1-1} \\ &=4 \cdot 3^{0} \\ &=4 \times 1=4 \\ T_{2} &=4 \cdot 3^{2-1} \\ &=4 \cdot 3^{1} \\ &=4 \times 3 \\ &=12 \\ T_{3} &=4 \cdot 3^{3-1} \\ &=4 \cdot 3^{2} \\ &=4 \times 9 \\ &=36 \end{aligned}$

$\begin{aligned} T_{4} &=4.3^{4-1} \\ &=4.3^{3} \\ &=4 \times 27 \\ &=108 \\ T_{5} &=4.3^{5-1} \\ &=4.3^{4} \\ &=4 \times 81 \\ &=324 \end{aligned}$

Hence, the terms are 4, 12, 36 , 108 , 324.

(ii) 

$\begin{aligned} T_{h} &=\frac{5^{h-1}}{2^{h+1}} \\ T_{1} &=\frac{5^{1-1}}{2^{n+1}} \\ &=\frac{5^{0}}{2^{2}} \\ &=\frac{1}{4} \\ T_{2} &=\frac{5^{2-1}}{2^{2+1}} \\ &=\frac{5^{1}}{2^{3}} \\ &=\frac{5}{8} \\ T_{3} &=\frac{5^{3-1}}{2^{3+1}} \\ &=\frac{5^{2}}{2^{4}} \\ &=\frac{25}{16} \end{aligned}$

$\begin{aligned} T_{4} &=\frac{5^{4-1}}{5^{9+1}} \\ &=\frac{5^{3}}{2^{5}} \\ &=\frac{125}{2^{5}} \\ &=\frac{125}{32} \\ T_{5} &=\frac{5^{5-1}}{2^{5+1}} \\ &=\frac{5^{4}}{2^{6}} \\ &=\frac{625}{64} . \end{aligned}$

Hence, the 5 terms are  $\frac{1}{4}, \frac{5}{8}, \frac{25}{16} ,\frac{125}{32}, \frac{625}{64}$

Question 5

Ans: (i) 12 , -36 .... sixth term

$\begin{aligned} \therefore \quad a &=12, \\ r &=\frac{-36}{12} \\ r &=-3 . \end{aligned}$

$\begin{aligned} \therefore T_{6} &=arn^{-1}  \\ &=12 \times(-3)^{6-1} \\ &=12 \times(-3)^{5} \\ &=12 \times(-243) \\ &=-2916 \end{aligned}$

(ii)  $3,-\frac{1}{3}, \ldots, 8$ th term.

$\begin{aligned} \therefore \quad q &=3, \\ r &=-\frac{1}{3} \div 3 . \\ &=-\frac{1}{3} \times \frac{1}{3} \\ &=-\frac{1}{9} \end{aligned}$

So , $\begin{aligned} T_{8} &=arn^{-1} \\ &=3 \cdot\left(-\frac{1}{9}\right)^{81} \\ &=3\left(-\frac{1}{9}\right)^{7} \end{aligned}$

(iii)

 $\begin{aligned} & b^{2} c^{3}, b^{3} c^{2}, \ldots, & 5th term \\ & \therefore a=b^{2} c^{3} \end{aligned}$

$r=\frac{b^{2} c^{2}}{b^{2} c^{8}}$

$r=\frac{b}{c}$

So $T_{5}= arn^{-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{5-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{4} .$

$=b^{2} c^{3} \times \frac{b^{4}}{c^{4}}$

$=b^{2} c^{2} \times b^{4} \times c^{-4}$

$=b^{2+4} \times c^{3}-4$

$=b^{6} \times c^{-1}$

$\frac{b^{6}}{c}$

Question 6

Ans: G.P. is $27,-18,12,-8, \ldots$ is $\frac{1024}{2187}$ (Given)

$\begin{aligned} \therefore a &=27, \\ r &=\frac{-18}{27} \\ r &=\frac{-2}{3} . \end{aligned}$

Let  $\frac{1024}{2187}$ be the nth term 

$\therefore a_{n}=a r n^{-1}$

$\frac{1024}{2187}=27\left(-\frac{2}{3}\right)^{n-1}$

$\frac{1024}{2187 \times 27}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{7} \times 3^{3}}=\left(-\frac{2}{3}\right)^{n-1}$

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

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

On comparing, 

$10=n-1$

$10+1=n$

$11=n$

n=11

Hence, it is 11th term.

 

Question 7

 

Ans: In a G.P 

$T_{4}=54$

$T_{7}=1458$

Let a be the first term and r be the common ratio

$So T_{4}=a r^{n-1}$

$54=ar^{4-1}$

$54=ar^{3}$

$a r^{3}=5 4$.........(i)

Dividing ,eqn (i) and (ii)

$r^{3}=\frac{1458}{54}$

$r^{3}=27$

$r=\sqrt[3]{27}$

$r=3$

Put the value of r = 3 in eq (i)

$a r^{3}=54$

$a \times(3)^{3}=54$

$a \times 27=54$

$a=\frac{54}{27} .$

$a=2$

$\therefore a=2, r=3 .$

Hence , G.P will be 2, 6 ,18 ,54....

 

Question 8

Ans: In a G.P. $3,3 \sqrt{3}, 9, \ldots$

Last term (l) = 2187

∴ a = 3

r =  $\frac{3 \sqrt{3}}{3}$

$T_{n}=l=arn^{-1}$

$2187=3(\sqrt{3})^{n-1}$

$\frac{2187}{3}=(\sqrt{3})^{n-1}$

$729=(\sqrt{3})^{n-1}$

$3^{6}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{6 \times 2}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{12}=(\sqrt{3})^{n-1} .$

On comparing,

$12=n-1$

$12+1=n$

$13=n$

$n=13$

Hence, it is 13th term

Question 9

Ans: Given , 

1, x, y , z 16 are in G.P 

So, first term (a) = 1, 

Common ratio(r) = $\frac{x}{1}$

$T_{5}=16$

$\begin{aligned} \therefore T_{5} &=a r(h-1) . \\ 16 &=9 r(5-1) \\ 16 &=9 r^{4} \\ 16 &=1 \times r^{4} \\(2)^{4} &=r^{4} \end{aligned}$

On comparing, 

r= 3

 So, the common ratio is 2 

Then, 

$\begin{aligned} r=\frac{x}{1} &=\frac{2}{1} \\ x &=2 \end{aligned}$

Also , the common ratio

$\frac{16}{z}=2$

$\begin{aligned}&z=\frac{16}{2} \\&z=8\end{aligned}$

Also, the common ratio 

$\begin{aligned} \frac{2}{y} &=2 \\ \frac{y}{y} &=2 \\ y &=\frac{8}{2} \\ y &=4 . \end{aligned}$

According to the question 

$\therefore x+y+z$

$=2+4+8$

=14 

Question 10

 

Ans:  In a G.P 

$T_{3}=18$,

$T_{7}=3 \frac{5}{9}=\frac{32}{9}$

Let a be the first term and r be the common ratio 

∴ $T_{n}=a r^{n-1}$

$T_{3}=a r^{3-1}$

$18=a r^{2}$

$a r^{2}=18$ ........(i)

$T_{7}=a r^{7-1}$

$\frac{32}{9}=a r^{6}$

$a r^{6}=\frac{32}{9}$...........(ii)

On dividing , 

$\frac{ar^{6}}{ar^{2}}=\frac{32}{9\times 18}$

$r^{4}=\frac{16}{81}$

$r^{4}=\left(\frac{2}{3}\right)^{4}$

On comparing, 

$\begin{aligned} & r=\frac{2}{3} \\ \therefore \quad & a r^{2}=18 \end{aligned}$

$a \times\left(\frac{2}{3}\right)^{2}=18$

$a \times \frac{4}{9}=18$

$a=\frac{81}{2}$

Then , 

$T_{10}=ar^{9}$

$=\frac{81}{2} \times\left(\frac{2}{3}\right)^{9}$

$=\frac{81}{2} \times \frac{2^{9}}{3^{9}}$

$\begin{aligned} &=\frac{3^{4} \times 2^{9}}{2 \times 3^{9}} \\=& \frac{2^{9-1}}{3^{9-4}} \\=& \frac{2^{8}}{3^{5}} \\=& \frac{256}{243} \end{aligned}$

Question 11

 

Ans: Given, 

In a G.P 

$T_{5}=P$,

$T_{8}=Q$,

$T_{11}=S$

Let a be the first term and r be the common ratio 

$\begin{aligned} \therefore \quad T_{5} &=ar^{5-1} \\ &=ar^{4} \\ &=p \end{aligned}$

$\begin{aligned} T_{8} &=a_{r} 8-1 \\ &=a^{7} \\ &=a \end{aligned}$

$\begin{aligned} T_{11} &=a r^{11-1} \\ &=a r^{10} \end{aligned}$

=5

$\begin{aligned} Q^{2} &=\left(ar7^{2}\right)^{2} \\ &=a^{2} r^{14} . \end{aligned}$

and p $\times s$ = $a r^{4} \times a r^{16} .$

$=a^{2} r^{4}$

Hence , prove, 

$Q^{2}=P \times S$

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

  Exercise 9 B

Question 1 

Ans:(i) First 15 terms of the AP : 2, 5 , 8 , 11....

Here ,  $a=2, d=5-2=3, n=15$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$s_{15}=\frac{15}{2}[2 \times 2+(15-1) \times 3]$

$=\frac{15}{2}[4+42]$

$=\frac{15}{2} \times 46=345$

(ii) First 50 terms of the A.P -27 , -23 , -19....

Here, a = - 27 , d =-23 -(-27)

= -23 +27

= 4 

n= 50

So  $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times(-27)+(50-1) \times 4]$

$S_{50}=25[-54 \times 49 \times 4]$

$=25[-54+196]$

$=25 \times 142$

$=3550$

(iii) Ans : First 17 terms of the A.P: $=\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}, \ldots$

Here, $a=\frac{1}{5}, d=\frac{-2}{10}-\frac{1}{5}$

$\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}$

and $n=17$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{17}=\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}-8\right]$

$=\frac{17}{2}\left[\frac{2-90}{5}\right)$

$=\frac{17}{2} \times \frac{-38}{5}$

$=\frac{-323}{5}$

$=-64 \frac{3}{2}$

(iv) Ans: First 24 terms of AP : 0 , 6, 1 , 7 , 28 ......

Here a = 0.6 d = 1.7 , 0.6 = 1.1, n = 24 

So , $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{24}=\frac{24}{2}[2 \times 0.6+(24-1)(1.1)]$

$=12[1.2+23 \times 1.1]$

$=12[1.2+25.3] 84$

$=12 \times 26.5$

$=318.0$

Question 2 

Ans: (i) $\begin{aligned}=& 34+32+30+\ldots+2 \\ & Here a=34, d=32-34=-2,1=2\\ & a_{n}=a+(n-1) d \\ \Rightarrow & 2=34+(n-1) \times(-2) \\ \Rightarrow & 2-34=-2(n-1) \\ \Rightarrow & \frac{-32}{-2}=n-1 \end{aligned}$

$\Rightarrow n-1=16$

$n=16+1$

$n=17$

So , $s_{n}=\frac{n}{2}(a+l) .$

$=\frac{17}{2}(34+2)$

$=\frac{17}{2} \times 36=306$

(ii) $7+9-\frac{1}{2}+12+\ldots+67$

Here, $a=7, d=9 \frac{1}{2}-7=2 \frac{1}{2}=\frac{5}{2}, 1=67$

$U=\left(a_{n}\right)=a+(n-1) d$

$\Rightarrow 67=7+(n-1)\left(\frac{5}{2}\right)$

$67-7=\frac{5}{2}(n-1)$

$\Rightarrow \frac{60 \times 2}{5}$

$=n-1$

$\Rightarrow n-1$

$=24$

= 24n = 24 +1 

=25 

So, 

$S_{25}=\frac{n}{2}[a+1]=\frac{25}{2}[7+67]$

=$\frac{25}{2} \times 74$

$=925$

Question 3

Ans: In an AP

$d=-2, a=100,1=-10$

$l=\left(a_{n}\right)=a+(n-1) d$

$-10=100+(n-1)(-2)$

$(n-1)(-2)=-10-100=-110$

$n-1=\frac{-110}{-2}=55$

$\Rightarrow n=55+1$

=56

$s_{56}=\frac{n}{2}[a+1]=\frac{56}{2}[100-10]$

$=28 \times 90$

$=2580$

Question 4

Ans: $\begin{aligned} \Rightarrow & \text { AP is } 54,51,48, \ldots \text { and } S_{n}=513 \\ & \text { Here, } a=54, d=51-54 \\=&-3 \end{aligned}$

$\begin{aligned} & S_{n}=\frac{n}{2}[2 a+(n-1) d] \\ & 513=\frac{n}{2}[2 \times 54+(n-1) \times(-3)] \\ \Rightarrow & 513 \times 2 \\=& n[108-3 n+3] \\ \Rightarrow & 10260 \\=& 108 n-3 n^{2}+3 n \\ \Rightarrow & 3 n^{2}-11 n+1026=0 \\ \Rightarrow & n^{2}-37 n+342=0 \\ \Rightarrow & n^{2}-18 n-19 n+342=0 \end{aligned}$

$\left\{\begin{array}{l}\because 342=-18 \times(-19) \\ -37=-18-19\end{array}\right\}$

$\begin{aligned} & \Rightarrow n(n-18)-19(n-18)=0 \\ \Rightarrow &(n-18)(n-19)=0 \end{aligned}$

Either n - 18 =0 , then n = 18

Or n- 19 =0 , then n = 19

So, Number of terms = 18 or 19 

Question 5

 

Ans: $\Rightarrow S_{9}=72, d=5$

let a be the first term,

n=9

So, $S_{g}=\frac{n}{2}[2 a+(n-1) d]$

$72=\frac{9}{2}[2 a+(9-1) \times 5]$

$\frac{72 \times 2}{9}=2 a+40$

$\Rightarrow 16=2 a+90$

$2 a=16-40=$

$2 a=-24$

$a=\frac{-24}{2}$

$a=-12$

 So a = -12

$a_{10}=a+(n-1) d$

$=-12+(10-1) \times 5$

$=-12+45$

$=-33$

Question 6

Ans: In an AP, 

Sum of first 6 terms 

=42

$a_{10}: a_{30}=1: 3$

Let a be the first term and d be the common difference , then

$S_{6}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 a+(6-1) d]$

$42=3(2 a+5 d)=6 a+15 d$

$\Rightarrow 6 a+15 d=42$

$a_{10}: a_{30}=1: 3$

$\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}$

$3 a+27 d=a+29 d$

$3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d$

$\Rightarrow a=d$

From (i) 

$6 a+15 a=42 \Rightarrow 21 a=42$

$\Rightarrow a=\frac{42}{21}=2$

$So, a=2, d=2$

So first term = 2

$a_{13}= a+(n-1) d=2+(13-1) \times 2$

$=2+12 \times 2=2+24$

$=26$

Question 7

Ans:  In a AP 

$a_{13}=4 \times a_{3}$

$a_{5}=16$

Let a be the first term and d be the common difference 

 So ,$\operatorname{a}_{5}=a+(n-1) d$

$=a+(5-1) d=a+4 d$

 So , $a+4 d=16$.............(i)

Similarly  

$a_{13}=a+12 d$ and $a_{3}=a+2 d$

$So, a+12 d=4 \times(a+2 d)$

$a+12 d=4 a+8 d$

$12 d-8 d=4 a-a \Rightarrow 3 a=4 d$

$a=\frac{4}{3} d$

From (i)

 $\frac{4}{3} d+4 d=16 \Rightarrow \frac{16}{3} d=16$

$\Rightarrow d=\frac{16 \times 3}{16}=3$

So, d= 3

and a = $\frac{4}{3} d=\frac{4}{3} \times 3=4$

$a=4, d=3$

Now $_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 4+(10-1) \times 3]$

$=5(8+27)$

$=5 \times 35=175$

Question 8

Ans:  $\Rightarrow A P$ is $8,10,12, \ldots$.

Were $a=8, d=10-8=2, n=60$

So  $T_{n}=a+(n-1) d$

$\Rightarrow 60=8+(n-1) \times 2$

$\Rightarrow(n-1) \times 2=60-8$

$=52$

$n-1=\frac{52}{2}$

$=26$

$n=26+1$

$=27$

$T_{60}=a+(60-1) d$

$=8+59 \times 2$

$=8+61=69$

Sum of last 10 term $=s_{60}-s_{50}$

$=\frac{60}{2}(2 a+59 d)-\frac{50}{2}(2 a+49 d)$

$=30(2 a+59 d)-25(2 a+49 d)$

$=60 a+1770 d-50 a-1225 d$

$=10 a+545 d=10 \times 8+545 \times 2$

$=80+1090$

$=1170$

Question 9

Ans: $\Rightarrow$ In an $A P$,

$\begin{aligned}&a_{12} 0 \times T_{12}=-13 \\&s_{4}=24\end{aligned}$

Let a be the first term and $b$ e the common difference, then

$a_{12}=a+(n-1) d$

$\Rightarrow-13=a+(12-1) d$

$\Rightarrow a+11 d=-13 \Rightarrow a=-13-11 d$

$s_{4}=\frac{n}{2}[2 a+(n-1) d]$

$24=\frac{4}{2}[2 a+3 d]=2[2 \times(-13-11 d)+3 d]$

$\frac{24}{2}=-26-22 d+3 d \Rightarrow 12=-26-19 d$

$\Rightarrow 12+26=-19 d \Rightarrow-19 d=38$

$d=\frac{38}{-19}=-2$

and $a=-13-11 d=-13+11 \times 2$

$=-13+22=9$

so, $a=9, d=-2$

Now , $s_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 9+(10-1)(-2)]$

$\begin{aligned} & 5[18+9(-2)]=5(18-18) \\=& 5 \times 0 . \\=& 0 \end{aligned}$

Question 10

Ans: $\because \Rightarrow$ Natural numbos betuteen 101 and 999 which are divisible by 2&5 both are 110 , $120,130, \ldots, 990$

Here $a,=110$ and $d=10,1=990$

 Now, $l=a_{n}=a+(n-1) d$ 

$990=110+(n-1) \times 10$

$\Rightarrow 990-110=10(n-1) \Rightarrow 10(n-1)=880$

$\Rightarrow n-1=\frac{880}{10} \Rightarrow n-1=88$

So , n = 88+ 1 = 89

Now , $s_{8 9}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{89}{2}[2 \times 110+(89-1) \times 10]$

$=\frac{89}{2}[220+88 \times 10]=\frac{89}{2}[220+880]$

$=\frac{89}{2} \times 110$

$=48950$

Question 11

Ans:  $\Rightarrow 2$ - digit number greater than 50 which are divisible by 7 , leaves a remainder of 4 are:

$53,60,67,74,81,88,95$

Here $a=53$,

$\begin{aligned} d &=7 \\ l &=95 \end{aligned}$

$\begin{aligned} \therefore a_{n}(l) &=a+(h-1) d . \\ 95 &=53+(h-1) \times 7 . \end{aligned}$

$\begin{aligned} 95-53 &=7(n-1) . \\ 42 &=7(n-1) \\ \frac{42}{7} &=(n-1) \\ 6 &=(n-1) \\(n-1) &=6 \\ n &=6+1 \\ n &=7 . \end{aligned}$

Then , 

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{7}{2}[2 \times 53+(7-1) \times 7]$

$=\frac{7}{2}[106+6 \times 7]$

$=\frac{7}{2}[106+42]$

$=\frac{7}{2} \times 48$

$=518$

Question 12

Ans: (i) Integers between 100 and 200 which are divisible by 9 are 108, 117 , 126 ,......198

Here 

a= 108

d =9

and l = 198

$\begin{aligned} \therefore a_{n}(l) &=a+(n-1) d . \\ 198 &=108+(n-1) \times 9 . \\ 198-108 &=9(n-1) \\ 90 &=9(n-1) \\ \frac{90}{9} &=n-1 \\ 10 &=n-1 \\ 10+1 &=n \\ n &=11 \\  & \end{aligned}$

Then 

$\begin{aligned} S_{11} &=\frac{h}{2}[2 a+(h-1) d] . \\ &=\frac{11}{2}[2 \times 108+(11-1) \times 9] \\ &=\frac{11}{2}[216+10 \times 9] \\ &=\frac{11}{2}[216+90] \end{aligned}$

$\frac{11}{2}\times 306$

=1683

(ii) Then sum of integers from 101 to 199

Here, a = 101,

d= 1 

l = 199

and n = 99

$s_{n}=$ $\frac{n}{2}[2 a+(n-1)d]$

$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$

$=\frac{99}{2}[202+98]$

$=\frac{99}{2}\times 300$

=14850

∴ Sum of integers which are not divisible by 9 

$=14850-1683$

$=13167$

Question 13

Ans:  Number between 9 and 95 when divided by 3, 

leaves remainder 1 

$10,13,16,19, \ldots . .99$

Here, 

$\begin{aligned} q &=10, \\ d &=3, \\ \text { and } l &=94 . \end{aligned}$

$a_{n}(l)=a+(n-1) d $

$94=10+(n-1) \times 3$

$94-10=3(n-1) .$

$84=3(n-1)$

$\frac{84}{3}=n-1$

$28=n-1$

$28+1=n$

$29=$n

$n=29$

Then, middle term 

$\begin{aligned} &=\frac{29+1}{2} \text { th } \\ &=\frac{30}{2} \mathrm{th} . \end{aligned}$

=15th term

Sum of first 14 terms 

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{74}{2}[2 \times 10+(14-1) \times 3]$

$=7[20+13 \times 3]$

$=7[20+39]$

$=7 \times 59$

$=413$

Middle terms = 10+15\times 3

= 10+45 = 55

After middle term; number of terms = 14 

Whose first terms (a)= 55 , d = 3

and n = 14

$\therefore s_{14}=\frac{19}{2}[2 \times 55+(14-1) \times 3]$

$\begin{aligned} &=7[110+39] \\ &=7 \times 149 \\ &=1043 \end{aligned}$

Question 14

Ans: $S_{n}=$ Sum of first n terms of an A.P.

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$To $ prove $=S_{12}=3\left(S_{8}-S_{4}\right)$

from $R \cdot H \cdot S=3\left[S_{8}-S_{4}\right]$

$=3\left[\frac{8}{2}[2 a+(8-1) d]-\frac{4}{2}[2 a+(4-1) d]\right] .$

$=3\left[\frac{8}{2}(2 a+7 d)-\frac{4}{2}[2 a+3 d)\right]$

$=3[4(2 a+7 d)-2(24+3 d)]$

$=3[84+28 d-4 a-6 d]$

$=3[4 a+22 d]$

$=12 a+66 d .$

$=6[2 a+11 d) .$

$\frac{12}{2}[2 a+(12-1) d]$= $S_{12}$ =L.H.S

Hence proved 

Question 15

Ans: Sum of first n even natural number $=\left(1+\frac{1}{h}\right)$ 

(Sum of first n odd natural numbers)

Sum of first n even natural number (2,4,6,8......2n)

$=\frac{n}{2}[2 a+(n-1) d] .$

$=\frac{n}{2}[2 \times 2+(n-1) \times 2]$

$=\frac{n}{2}[4+2n-2] .$

$=\frac{n}{2}[2+2n] .$

$=\frac{n}{2} \times 2(1+n)$

$=n(1+n) $

and Sum of n odd natural number (1, 3,5 ,.......(2n -1))

$=\frac{n}{2}[2 \times 1+(n-1) 2]$

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}[2 n]$

$=\frac{n}{2} \times 2[n]$

$=n^{2} .$

Then , 

$n^{2} \times \left(1+\frac{1}{n}\right)$

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

$=n^{2} \times \frac{n+1}{n}$

$=n(n+1)$

Question 16

Ans: Given,

First term $\left(a_{1}\right)=5$

Common difference $\left(d_{1}\right)=36$.

$S_{n}=$ Sum of $2_{n}$ terms of another $A P$ whose first term $\left(a_{2}\right)=36$ and common difference $\left(d_{2}\right)=5$.

 In first $A P$

$\begin{aligned} S_{n} &=\frac{n}{2}[2 a+(n-1) d] . \\ &=\frac{n}{2}[2 \times 5+(n-1) \times 36] \\ &=\frac{n}{2}[10+36 n-36] \\ &=\frac{n}{2}[36 n-26] . \end{aligned}$

$=\frac{n}{2} \times 2[18 n-13]$

$=n[18 n-13]$

In Second AP, 

$S_{2 N}=\frac{2 n}{2}[2 \times 36+(2 n-1) \times 5]$

$=n[72+10{n}-5]$

$=\mathrm{n}[67+10 \mathrm{n}] $

$\therefore S_{n}=S_{2 n}$

18 (18n- 13) = n(67 + 10n)

18 n - 13 =  $\frac{n}{n}(67+10 n)$

$18 n-10 n=67+13$

8n = 80

n = $\frac{80}{8}$

n=10

Question 17

Ans: Sum of first 10 terms of an AP = $4 \times sum$ of first 5 terms 

Let a be the first term and d be the common difference 

$\begin{aligned} \therefore S_{n} &=\frac{n}{2}[2 a+(n-1) d] \\ S_{10} &=\frac{10}{2}[2 a+(10-1) d] \\ &=5[2 a+9 d] \\ &=10a+45d \end{aligned}$

$\begin{aligned} S_{{5}} &=\frac{5}{2}[2 a+(5-1) d] \\ &=\frac{5}{2}[2 a+4 d] \end{aligned}$

$=\frac{5}{2} \times 2[a+2 d]$

$=5 a+10 d$

According to the question,

10a + 45 d =  $4 \times(5 a+10 d)$

$10 a+45 d=20 a+40 d .$

$45 d-40 b=20 a-10 a$

$5 a=109 .$

$\frac{5}{10}=\frac{a}{d}$

$\frac{a}{d}=\frac{5}{10}$

$\frac{a}{d}=\frac{1}{2}$

∴ Ratio in first term to the common difference = 1:2

Question 18

Ans:  Given, 

In the first row, plants are = 37

In second row = 35 

In third row = 33

and in the last row = 5

Here , 

a= 37 

d= 35 - 37 

d= - 2

$\begin{aligned} a_{n}(1)=& a+(n-1) d . \\ 5=& 37+(n-1) \times(-2) \\ 5 &-37=-2 n+2  \\ &-32=-2(n-1) . \\ & \frac{+32}{+2}=n-1 \end{aligned}$

$16=n-1$

$16+1=n$

$17=n$

$n=17$

So, number of rows = 17 

Then, 

Total plants $\left(S_{n}\right)=\frac{n}{2}[24+(n-1) d]$

$=\frac{17}{2}[2 \times 37+(17-1) \times(-2)] .$

$=\frac{17}{2}[74-32]$

=357

Hence , the total plants is 357

Question 19

Ans: Given , 

Total logs = 200 

In the bottom row, number of logs = 20 

and next row above it = 19 

Next row above it = 18

And Soon 

- AP is 20,19,18,17,16....... and $S_{n}=200$

Let the top row be the nth row 

$\therefore a_{n}=a+(n-1)d$

and $S_{n}=200$

So, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2 \times 20+(n-1) \times(-1)]$

$400=n[40-n+1]$

$400=40 n-n^{2}+h$

$400=41 n-n^{2} $

$n^{2}-41 n+400=0$

$n^{2}-25 n-16 n+400=0$

$n(n-25)-16(n-25)=0$

$(n-25)(n-16)=0$

n -25 =0 or n-16=0

n=25 or n =16

But n = 25 is not possible 

 As number of logs in the first row = 20 

and d = - 1 

Number of row = 16 

and number of log in 16th row 

$=9+(n-1) d$

$=20+(16-1) \times(-1)$

$=20-15$

$=5 logs$

Question 20

Ans: Giver,

Total amount= ₹1590,

Number of cash prizes = 7 

and each prize is Rs 50 less than the proceeding prize 

Let first prize = Rs a 

Then second prize = a - 50 

Third prize = a - 100

And so on 

$\therefore$ first term $=a$,

common difference (d) $=-50$,

$S_{n}=1890$

$n=7 $

$\begin{aligned} S_{n}=& \frac{n}{2}[2 a+(n-1) d] \\ 1890=& \frac{7}{2}[2 a+(7-1) \times(-50)] \\ 3780=& 7[2 a-300] \\ & \frac{3780}{7}=2 a-300 \\ 540=& 2 a-300 \\ 540+300=& 2 a \\ 840=& 2 a \\ \frac{840}{2}=9 \\ 420=a \\ a=420 \end{aligned}$

Hence prizes are $ ₹420, ₹ 370, ₹ 320 ; 270, ₹ 220, ₹ 170$ and $₹ 120$.

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

 Exercise 9 A

Question 1 

Ans: (i) Yes 

(ii) Yes, it is on arithmetic progression as cost of digging in the beginning is Rs 125 for the first meter and then increasing Rs 65 for every subsequent meter. 

Question 2

Ans: (i) a= 10 , d= 7 

10 , 17 , 24 ,31 , 38 

(ii) a = 55 , d = -4

So AP =55, 51 47, 43

(iii) a =-1 , d=  $\frac{1}{2}$

$A P=-1, \frac{-1}{2}, 0, \frac{1}{2}, 1$

(iv) $a=-3: 25, d=-0.25$

$A P=-3.25,-3.50,-3.75,-4.90,-4.25$

Question 3

Ans:  $\Rightarrow$ List of numbers : $-12,-9,-6,-3,0,3$

 Here $a=-12, d=-9-(-12)=-9+12=3$ 

$a=-12, d=3$

(ii)  $\Rightarrow$ In an AP, a $=3, d=0, n=7$,

$\therefore a_{n}=a+(n-1) d=3+(7-1)+0$

$=3+0=3$

Question 4

Ans: (i) $\Rightarrow A P=-5,-\frac{5}{2}, 0, \frac{5}{2}, \ldots$

$\therefore, a=-5, d=\frac{-5}{2}-(-5)$

$=\frac{-5}{2}+5=\frac{5}{2}$

$So , a_{25}=a+(n-1) d=-5+(25-1) \times \frac{5}{2}$

$=-5+24 \times \frac{5}{2}$

$=-5+60$

$=55$

(ii) AP is 27, 22, 17 ....

∴ a = 27 , d = (22- 27 ) = -5

So $a_{30}=a+(n-1) d$

$=27+(30-1) \times(-5) .$

$=27+29(-5)$

$=27-145=-118$

(ii) $\Rightarrow A P=-0.1,-0.2,-0.3, \ldots$

$\therefore a=-0.1$

$d=-0.2-(-0.1)=-0.2+0.1=-0.1$

$So , a_{10}=a+(n-1) d=-0.1+(10-1)(-0.1)$

$=-0.1+9 \times(-0.1)=-0.1-0.9=-1$

Question 5

Ans:  K , 2 k - 1 & 2K + 1 are the three Terms of an AP, then 

$a=k, d=2 k-1-k=k-1$ (i)

and $d=2 k+1-(2 k-1)=2 k+1-2 k+1$

$=2$ (ii)

From (i) and (ii)
So , K-1 = 2 = k = 2+1 = 3

SO k = 3 

Question 6

Ans: $\Rightarrow A p$ is $\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}$

$d=\frac{1-6 q}{3 q}-\frac{1}{3 q}=\frac{1-6 q-1}{3 q}=\frac{-6 q}{3 q}$

$=-2$

So , Common difference = -2

Question 7

Ans: = Which term of Ap 5, 13 , 21 , ...... is 181 

Let it be nth term, then 

In AP a = 5, d = 13 -5 = 8

$a_{n}=a+(n-1) d$

$181=5+(n-1) \times 8$

$\Rightarrow|8|=5+8 n-8$

$\Rightarrow 8 n=181-5+8=184$

$\Rightarrow n=\frac{184}{8}=23$

so, 181 is 23 rd term.

Question 8

Ans:

 $\begin{aligned}&\Rightarrow \text { Ap is } 3,10,17, \ldots \\&\therefore a=3, d=10-3=7\end{aligned}$

Let nth term is 84 more than its 13 the term

$\begin{aligned}&a_{13}=a+(n-1) d=3+(13-1) \times 3 \\&=3+12 \times 7=87 \\&a_{n}=3+(n-1) \times 7=3+7 n-7=7 n-4 \\&\text { Now, } 7 n-4=84+87=171 \\&7 n=171+4=175 \\&n=\frac{175}{7} \\&=25\end{aligned}$

So, 25 th is the required term.

Question 9

Ans: In an AP 

6th term = 19 and 

16th term = 15 + 11th term 

Let a be the first term and d be the common difference 

So , $a_{6}=a+(n-1) d$

$\Rightarrow 19=a+5 d$.......(i)

$\Rightarrow a+5 d=19$

similarly

$a_{11}=a+10 d$ and $a_{16}=a+15 d$

Now, $a+15 d=a+10 d+15$

$15 d-10 d=15$

$\Rightarrow 5 d=15$

$\Rightarrow d=\frac{15}{5}=3$

Erom (i)

$a+5 \times 3=19$

$\Rightarrow a+15=19$

$\Rightarrow a=19-15=4$

$or=4, d=3$

So, $A P$ is $4,7,10,13,16 \ldots$

Question 10

Ans:   $\Rightarrow$ 2nd term $+7$ th term $=30$

15 th term $=2 \times 8$ th term $-1$

let a be the first term and $d$ be the common difference, then

$a_{2}=a+(n-1) d=a+(2-1) d=a+d$

similarly, $a=a+(7-1)$

$d \Rightarrow a+6 d$

$a_{15}=a+(15-1)$

$d \Rightarrow a+14 d$ and

$a_{8}=a+(8-1) d$

$\Rightarrow a+7 d$

Now, $a_{2}+a_{7}=30$

$\Rightarrow a+d+a+6 d=30$

$\Rightarrow 2 a+7 d=30$

and $a+14 d=2 \times(a+7 d)-1$

$a+14 d=2 a+14 d-1$

$2 a-a=1$

$\Rightarrow a=1$

From (i), $2 \times 1+7 d=30$

$\Rightarrow 2+7 d=30$

$\Rightarrow 7 d=30-2=28$

$\Rightarrow d=\frac{28}{7} \Rightarrow=4$

So , d = 4 a = 1

So , AP  = 1, 5 ,9 , 13 , 17.....

Question 11

Ans:  In an Ap 

4th term $(a 4)$

=11

5 th term $+7$ th iterm

$=34$

let a be the first term and $d$ be the common difference, then

$\begin{aligned} & a_{4} \\=& a+(n-1) d \\=& a+(4-1) d \\ \Rightarrow & a+3 d=11 \end{aligned}$

similarly.

$a_{7}=a+(7-1) d \Rightarrow a+6 d$

We know that, $a_{4}+a_{7}=34$

$\therefore a+4 d+a+6 d=34$

$\Rightarrow 2 a+10 d=34$

$a+5 d=17$

subtracting (ii) from (i).

$\begin{aligned}&=(a+3 d)-(a+5 d)=11-17 \\&=a+3 d-a-5 d=-6 \\&-2 d=-6\end{aligned}$

$2 d=6 \Rightarrow d=\frac{6}{2}=3$

$\therefore$ Common difference $=3$

Question 12

Ans: $\Rightarrow A P=213,205,197, \ldots .37$

let 37 be the $n$th term

Now, $a=213 \mathrm{~A} d=-205-213=-8$

$\therefore a_{n}=a+|n \cdot 1| d$

$\Rightarrow 37=213+(n-1)(-8)$

$37=213-8 n+8$

$n=\frac{184}{8}=23$

There are 23 term in the AP 

Middle term = $\frac{23+1}{2}=12$ th term

$a_{12}=a+11 d$

$2213+11(-8)=213-88=125$

Question 13 

Ans: In an AP 

3rd term = 4

9th term =- 8 

Which term is 0 

Let a be the first term and d be the common difference, then 

$\begin{array}{rl} & a_{3}=a+(n-1) d \\ \Rightarrow & 4=a+(3-1) d \\ \Rightarrow & a+2 d=4 \\ \text { similarly, } a+8 d=-8 \\ \text { subtracting (i) from (ii), } \\ \Rightarrow(a+8 d)-(a+2 d)=-8-4 \\ 6 & d=-12 \Rightarrow d=\frac{-12}{6}\end{array}$

=-2

and $a+2(-2)=4$

$\Rightarrow a-4=4$

$\Rightarrow a=4+4=8$

$\therefore a=8, d=-2$

Let $a_{n}$ be equal to 0 , then

$\begin{aligned} & a+(n-1) d=0 \\ \Rightarrow & 8+(n-1)(-2)=0 \\ \Rightarrow &+8-2 n+2=0 \Rightarrow 10-2 n=0 \end{aligned}$

$\Rightarrow 2 n= 10$

$n=\frac{10}{x}$

$=5$

=5th  term is zero

Question 14

Ans: $\Rightarrow$ In an AP

8 th $\operatorname{tern}\left(a_{8}\right)=0$

To prove that 38 th term

$=3 \times 18$ th term

$=$ Let $a$ be the first teron and $d$ be the common

$=$ difference, then

$\begin{aligned}&a_{8}=a+(n-1) d \\&\Rightarrow a+(8-1) d=0 \\&\Rightarrow a+7 d=0 \\&\Rightarrow a=-7 d\end{aligned}$

$\begin{aligned}&\text { Similarly } \\&a_{18}=a+17 d \\&a_{38}=a+37 d\end{aligned}$

Now , a + Ad= -7d + 17d 

 = 10 d  and a + 37 d 

= - 7d + 37 d 

= 30 d 

So , 30 d 

$=3 \times 10 \mathrm{~d}$

$=30 \mathrm{~d}$

$= a_{38}=3 \times a_{18}$

Question 15

Ans: AP is 120 , 116 , 112.....

Which first term of this Ap will be negative 

Here , a =120, d = 116 - 120 =-4 

Let nth term of the given AP be the negative term  $a_{n}$ or $T_{n}<0$

$\Rightarrow a+(n-1) d<0$

$\Rightarrow 120+(n-y)(-4)<0$

$\Rightarrow 120-4 n+4<0$

$\Rightarrow 124-4 n<0 .$

$\Rightarrow 4 n>124$

$\Rightarrow n>\frac{124}{4} $

= 31

So', $n=32$ be the first negative term

$\Rightarrow 32$ nd term is the first negative term

Question 16

Ans: $\Rightarrow 3$-digits terms are $100,101,102, \ldots, 999$ and terms which are divisible by 7 Will be $105,112,119, \ldots, 994$

$\because a=105, d=7$

Last term ($a_{n}$) = 994

$a_{n}=a+(n-1) d$

$994=105+(n-1) \times 7$

$=105+7 n-7=7 n+98$

$7 n=994-98$

$=896$

$n=\frac{896}{7}=128$

Question 17

Ans:- Let $d$ be the common difference of the two $A P$ series $\alpha_{1}$ and $a_{2}$ are the first term of the two AP's respectively.

In first $A P, a_{1}=1, a_{2}=-8$

Difference between their 4 th terms

$=\left(a_{1}+3 d\right)-\left(a_{2}+3 d\right)-$

$=-1-(-8)$

$a_{1}+3 d-a_{2}-3 d$

$=-1+8$

$\Rightarrow a_{1}-a_{2}=7$

$or a_{2}+3 d-a_{1}-3 d$

$=-8-(-1)$

$a_{2}-a_{1}=-8+1=-7$

$\therefore$ Difference $=7$ or $-7$

Question 18

Ans:  $\Rightarrow$ In an AP

Ratio between 4 th term and 9 th term $=1: 3$

let a be the first term and $d$ be the common difference, then

4 th term $\left(a_{4}\right)$

$=a+(n-1) d$

$=a+(4-1) d$

$=a+3 d$

Similarly 9th term = a + 8 d 

Now,

So, $\frac{a+3 d}{a+8 d}=\frac{1}{3}$

$\Rightarrow 3 a+9 d=a+8 d$

$\Rightarrow 3 a-a$

$=3 d-9 d$

$\Rightarrow 2 a=-d$

$\Rightarrow d=-2 a$

Now, $a_{12}=a+(12-1) d$

$=a+11 d$

and $a_{5}=a+(5-1) d$

$=a+4 d$

So,$ \frac{a+11 d}{a+4 d}$

$\frac{a+(-2 a) \times 11}{a+4(-2 a)}$

$=\frac{a-22 a}{a}-8 a$

$=\frac{-21 a}{-7 a}$

$=\frac{3}{1}$

$=3: 1$

Question 19

Ans: 7th term from the end of AP 

7, 10 , 13, ......184 

Hence , a = 7, d =10-7 = 3

$\begin{aligned} & T_{n}(1)=184 \\ T_{n}=a+(n-1) d \\ \Rightarrow & 7+(n-1) \times 3 \\ \Rightarrow & 7+3 n-3 \\=& 4+3 n \end{aligned}$

$\operatorname{So} 4+3 n=184$

$\Rightarrow 3 n=184-4$

$=180$

$n=\frac{180}{3}=60$

7th term from the last will be 60- (7-1)

60 - 6 = 54th from the beginning 

So $T_{54}=a+53 d$ 

$\begin{array}{rl} & =7+53 \times 3 \\  & 7+159 \\ = & 166\end{array}$

Question 20

Ans: Four angles of a quadrilateral are in AP 

Sum of the first 3 angles = 2 $\times$ 4th angle

Adding 4th angle to both sides 

Sum of 4 angle $=2 \times 4th angle +4th angle

$=3 \times 4$th angles

But Sum of 4 angles of a quadrilateral = 360 

So,  $=3 \times 4$th angles

4 th angle $=\frac{360^{\circ}}{3}=120^{\circ}$

$\Rightarrow 4$ th angle $=120^{\circ}$

Let a be the first term (angle) and d be the common difference 

So, a + 3d = 120 .............(i)

And $a+a+d+a+2 d=360^{\circ}-120^{\circ}=240$

$3 a+3 d=240$..............(ii)

Subtracting 2a= 120 = $a=\frac{120^{\circ}}{2}=60^{\circ}$

But $a+3 d=120$

$\Rightarrow 60+3 d= 120 \Rightarrow 3 d$

$=120-60=60$

$d=\frac{60^{\circ}}{3}=20^{\circ}$

So, Angles are  $60^{\circ}, 80^{\circ}, 100^{\circ}, 120^{-}$

Question 21

Ans: Sum of 3 numbers in $A P=-3$ 

and product $=8$ 

let three numbers in AP be 

$a-d, a, a+d$

$a-d+a+a+d$

=-3

$\Rightarrow 3 a=-3$

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

$=-1$

and (a- d) $a(a+d)=8$

$\left(a^{2}-d^{2}\right) a=8$

$\left.\left[(-1)^{2}-d^{2}\right)\right](-1)=8$

$1-d^{2}=-8$

$d^{2}=1+8$

$d_{2}=9$

$d^{2}=(+3)^{2}$

So, Number be -1 , -3 ,-1 -1, +3

= -4 , -1 , 2 

Or -1+3, -1, -1 , -3

= 2 , -1 , -4 

Question 22

Ans: Ram prasad's savings in first week = 10 rs 

Then his weekly saving is increasing = Rs 2.75 

So, a = Rs 10 and d = Rs 2.75 

nth week saving = rs 59.50

$a_{n}=a+(n-1) d$

$59.50=10+(n-1)(2.75)$

$59.50-10.00=(n-1)(2.75)$

$49.50=(n-1)(2.75)$

So, $n-1=\frac{49.50}{2.75}$

$=18$

so, $n=18+1$

$=19$

Question 23

 

Ans: $\Rightarrow$ Eer an AP

$T_{p}+T_{p}+2 q=2 T_{p+q}$

Let $a$ be the first term and $d$ be the common difference, then

$T_{p}=a+(p-1) d$,

$T_{p}+2 q=a+(p+2 q-1) d$

$T_{p}+q=a+(p+q-1) d$

Now, LHS $=T_{p}+T_{p+2 q}$

$=a+(p-1) d+a+(p+2 q-1) d$

$=a+d p-d+a+p d+2 q d-d$

$=2 a+2 d p+2 q d-2 d$

$=2(a+d p+q d-d)$

$=2[a+(p+q-1) d]$

$=2 \times T_{p+q}$

=R H S

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

  Exercise 9D

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

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

  Exercise 9C

Question 1 

Ans: (i) $27,9,3,1, \ldots$

$\begin{aligned}\therefore & a=27 \\r &=\frac{9}{27}=\frac{1}{3}, \frac{3}{9} \\&=\frac{1}{3}, .....\end{aligned}$

Hence, it is a G.P. and $r=\frac{1}{3}$

(ii) $-1,2,4,8, \ldots .$

$\therefore a=-1$,

$\quad b=\frac{2}{-1}=-2, \frac{4}{2} .$

$\quad=2, \frac{8}{4}=2 .$

So, it is not a G.P 

(iii) 

$\begin{aligned} & 2, \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \ldots \\ \therefore & a=2, \\ &=\frac{1}{2} \div 2 \\ &=\frac{1}{2} \times \frac{1}{2} \\ &=\frac{1}{4} . \\ \frac{1}{8} &=\frac{1}{2} . \end{aligned}$

$=\frac{1}{4} .$

$\frac{1}{32}=\frac{1}{8} .$

$=\frac{1}{32} \times 8 .$

$=\frac{1}{4}$

∴ IT is GP and  $r=\frac{1}{4}$

$(i v)-12,-6,0,6, \ldots . .$

$\begin{aligned}&\therefore a=-13 \\&r=\frac{-6}{-12}=\frac{1}{2} \\&=\frac{0}{-6} \\&=0\end{aligned}$

Hence,  it is not G.P.

Question 2

Ans:  (i) 2,6....

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

2,6,18,54,162

(ii) $\frac{1}{16},-\frac{1}{8}, \ldots$

$\left(r=-\frac{1}{8} \div \frac{1}{10}=-\frac{1}{8} \times 16=-2 .\right)$

$\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}, 1$

(iii)0.3, 0.06......

$r=\frac{0.01}{0.3}$= $\frac{1}{5}=0-2$

$0.3,0.03,0.012,0.0024,0.00048$

Question 3

Ans: 6th term of the G.P 2, 10 , 50......

∴ a = 2 

r=  $\frac{10}{2}=5$

$\begin{aligned} So  T_{6} &=9 x^{h-1} \\ &=2 \times 5^{6-1} \\ &=2 \times 5^{5} \\ &=2 \times 3125 \\ &=6250 \end{aligned}$

(ii)  11th term of the G.P. $4,12,3, \ldots .$

$\begin{aligned} \therefore & a=4 \\ & r=\frac{12}{4}=3 . \end{aligned}$

$\begin{aligned} T_{11} &=ar^{h-1} \\ &=4 \times(3)^{11-1} \\ &=4 \times 3^{10} \\ &=4 \times 59,049 \\ &=236196 \end{aligned}$

Question 4

Ans:

 (i) $\begin{aligned} T_{n} &=4 \cdot 3^{h-1} \\ \therefore T_{1} &=4 \cdot 3^{1-1} \\ &=4 \cdot 3^{0} \\ &=4 \times 1=4 \\ T_{2} &=4 \cdot 3^{2-1} \\ &=4 \cdot 3^{1} \\ &=4 \times 3 \\ &=12 \\ T_{3} &=4 \cdot 3^{3-1} \\ &=4 \cdot 3^{2} \\ &=4 \times 9 \\ &=36 \end{aligned}$

$\begin{aligned} T_{4} &=4.3^{4-1} \\ &=4.3^{3} \\ &=4 \times 27 \\ &=108 \\ T_{5} &=4.3^{5-1} \\ &=4.3^{4} \\ &=4 \times 81 \\ &=324 \end{aligned}$

Hence, the terms are 4, 12, 36 , 108 , 324.

(ii) 

$\begin{aligned} T_{h} &=\frac{5^{h-1}}{2^{h+1}} \\ T_{1} &=\frac{5^{1-1}}{2^{n+1}} \\ &=\frac{5^{0}}{2^{2}} \\ &=\frac{1}{4} \\ T_{2} &=\frac{5^{2-1}}{2^{2+1}} \\ &=\frac{5^{1}}{2^{3}} \\ &=\frac{5}{8} \\ T_{3} &=\frac{5^{3-1}}{2^{3+1}} \\ &=\frac{5^{2}}{2^{4}} \\ &=\frac{25}{16} \end{aligned}$

$\begin{aligned} T_{4} &=\frac{5^{4-1}}{5^{9+1}} \\ &=\frac{5^{3}}{2^{5}} \\ &=\frac{125}{2^{5}} \\ &=\frac{125}{32} \\ T_{5} &=\frac{5^{5-1}}{2^{5+1}} \\ &=\frac{5^{4}}{2^{6}} \\ &=\frac{625}{64} . \end{aligned}$

Hence, the 5 terms are  $\frac{1}{4}, \frac{5}{8}, \frac{25}{16} ,\frac{125}{32}, \frac{625}{64}$

Question 5

Ans: (i) 12 , -36 .... sixth term

$\begin{aligned} \therefore \quad a &=12, \\ r &=\frac{-36}{12} \\ r &=-3 . \end{aligned}$

$\begin{aligned} \therefore T_{6} &=arn^{-1}  \\ &=12 \times(-3)^{6-1} \\ &=12 \times(-3)^{5} \\ &=12 \times(-243) \\ &=-2916 \end{aligned}$

(ii)  $3,-\frac{1}{3}, \ldots, 8$ th term.

$\begin{aligned} \therefore \quad q &=3, \\ r &=-\frac{1}{3} \div 3 . \\ &=-\frac{1}{3} \times \frac{1}{3} \\ &=-\frac{1}{9} \end{aligned}$

So , $\begin{aligned} T_{8} &=arn^{-1} \\ &=3 \cdot\left(-\frac{1}{9}\right)^{81} \\ &=3\left(-\frac{1}{9}\right)^{7} \end{aligned}$

(iii)

 $\begin{aligned} & b^{2} c^{3}, b^{3} c^{2}, \ldots, & 5th term \\ & \therefore a=b^{2} c^{3} \end{aligned}$

$r=\frac{b^{2} c^{2}}{b^{2} c^{8}}$

$r=\frac{b}{c}$

So $T_{5}= arn^{-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{5-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{4} .$

$=b^{2} c^{3} \times \frac{b^{4}}{c^{4}}$

$=b^{2} c^{2} \times b^{4} \times c^{-4}$

$=b^{2+4} \times c^{3}-4$

$=b^{6} \times c^{-1}$

$\frac{b^{6}}{c}$

Question 6

Ans: G.P. is $27,-18,12,-8, \ldots$ is $\frac{1024}{2187}$ (Given)

$\begin{aligned} \therefore a &=27, \\ r &=\frac{-18}{27} \\ r &=\frac{-2}{3} . \end{aligned}$

Let  $\frac{1024}{2187}$ be the nth term 

$\therefore a_{n}=a r n^{-1}$

$\frac{1024}{2187}=27\left(-\frac{2}{3}\right)^{n-1}$

$\frac{1024}{2187 \times 27}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{7} \times 3^{3}}=\left(-\frac{2}{3}\right)^{n-1}$

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

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

On comparing, 

$10=n-1$

$10+1=n$

$11=n$

n=11

Hence, it is 11th term.

 

Question 7

 

Ans: In a G.P 

$T_{4}=54$

$T_{7}=1458$

Let a be the first term and r be the common ratio

$So T_{4}=a r^{n-1}$

$54=ar^{4-1}$

$54=ar^{3}$

$a r^{3}=5 4$.........(i)

Dividing ,eqn (i) and (ii)

$r^{3}=\frac{1458}{54}$

$r^{3}=27$

$r=\sqrt[3]{27}$

$r=3$

Put the value of r = 3 in eq (i)

$a r^{3}=54$

$a \times(3)^{3}=54$

$a \times 27=54$

$a=\frac{54}{27} .$

$a=2$

$\therefore a=2, r=3 .$

Hence , G.P will be 2, 6 ,18 ,54....

 

Question 8

Ans: In a G.P. $3,3 \sqrt{3}, 9, \ldots$

Last term (l) = 2187

∴ a = 3

r =  $\frac{3 \sqrt{3}}{3}$

$T_{n}=l=arn^{-1}$

$2187=3(\sqrt{3})^{n-1}$

$\frac{2187}{3}=(\sqrt{3})^{n-1}$

$729=(\sqrt{3})^{n-1}$

$3^{6}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{6 \times 2}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{12}=(\sqrt{3})^{n-1} .$

On comparing,

$12=n-1$

$12+1=n$

$13=n$

$n=13$

Hence, it is 13th term

Question 9

Ans: Given , 

1, x, y , z 16 are in G.P 

So, first term (a) = 1, 

Common ratio(r) = $\frac{x}{1}$

$T_{5}=16$

$\begin{aligned} \therefore T_{5} &=a r(h-1) . \\ 16 &=9 r(5-1) \\ 16 &=9 r^{4} \\ 16 &=1 \times r^{4} \\(2)^{4} &=r^{4} \end{aligned}$

On comparing, 

r= 3

 So, the common ratio is 2 

Then, 

$\begin{aligned} r=\frac{x}{1} &=\frac{2}{1} \\ x &=2 \end{aligned}$

Also , the common ratio

$\frac{16}{z}=2$

$\begin{aligned}&z=\frac{16}{2} \\&z=8\end{aligned}$

Also, the common ratio 

$\begin{aligned} \frac{2}{y} &=2 \\ \frac{y}{y} &=2 \\ y &=\frac{8}{2} \\ y &=4 . \end{aligned}$

According to the question 

$\therefore x+y+z$

$=2+4+8$

=14 

Question 10

 

Ans:  In a G.P 

$T_{3}=18$,

$T_{7}=3 \frac{5}{9}=\frac{32}{9}$

Let a be the first term and r be the common ratio 

∴ $T_{n}=a r^{n-1}$

$T_{3}=a r^{3-1}$

$18=a r^{2}$

$a r^{2}=18$ ........(i)

$T_{7}=a r^{7-1}$

$\frac{32}{9}=a r^{6}$

$a r^{6}=\frac{32}{9}$...........(ii)

On dividing , 

$\frac{ar^{6}}{ar^{2}}=\frac{32}{9\times 18}$

$r^{4}=\frac{16}{81}$

$r^{4}=\left(\frac{2}{3}\right)^{4}$

On comparing, 

$\begin{aligned} & r=\frac{2}{3} \\ \therefore \quad & a r^{2}=18 \end{aligned}$

$a \times\left(\frac{2}{3}\right)^{2}=18$

$a \times \frac{4}{9}=18$

$a=\frac{81}{2}$

Then , 

$T_{10}=ar^{9}$

$=\frac{81}{2} \times\left(\frac{2}{3}\right)^{9}$

$=\frac{81}{2} \times \frac{2^{9}}{3^{9}}$

$\begin{aligned} &=\frac{3^{4} \times 2^{9}}{2 \times 3^{9}} \\=& \frac{2^{9-1}}{3^{9-4}} \\=& \frac{2^{8}}{3^{5}} \\=& \frac{256}{243} \end{aligned}$

Question 11

 

Ans: Given, 

In a G.P 

$T_{5}=P$,

$T_{8}=Q$,

$T_{11}=S$

Let a be the first term and r be the common ratio 

$\begin{aligned} \therefore \quad T_{5} &=ar^{5-1} \\ &=ar^{4} \\ &=p \end{aligned}$

$\begin{aligned} T_{8} &=a_{r} 8-1 \\ &=a^{7} \\ &=a \end{aligned}$

$\begin{aligned} T_{11} &=a r^{11-1} \\ &=a r^{10} \end{aligned}$

=5

$\begin{aligned} Q^{2} &=\left(ar7^{2}\right)^{2} \\ &=a^{2} r^{14} . \end{aligned}$

and p $\times s$ = $a r^{4} \times a r^{16} .$

$=a^{2} r^{4}$

Hence , prove, 

$Q^{2}=P \times S$

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

  Exercise 9B

Question 1 

Ans:(i) First 15 terms of the AP : 2, 5 , 8 , 11....

Here ,  $a=2, d=5-2=3, n=15$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$s_{15}=\frac{15}{2}[2 \times 2+(15-1) \times 3]$

$=\frac{15}{2}[4+42]$

$=\frac{15}{2} \times 46=345$

(ii) First 50 terms of the A.P -27 , -23 , -19....

Here, a = - 27 , d =-23 -(-27)

= -23 +27

= 4 

n= 50

So  $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times(-27)+(50-1) \times 4]$

$S_{50}=25[-54 \times 49 \times 4]$

$=25[-54+196]$

$=25 \times 142$

$=3550$

(iii) Ans : First 17 terms of the A.P: $=\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}, \ldots$

Here, $a=\frac{1}{5}, d=\frac{-2}{10}-\frac{1}{5}$

$\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}$

and $n=17$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{17}=\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}-8\right]$

$=\frac{17}{2}\left[\frac{2-90}{5}\right)$

$=\frac{17}{2} \times \frac{-38}{5}$

$=\frac{-323}{5}$

$=-64 \frac{3}{2}$

(iv) Ans: First 24 terms of AP : 0 , 6, 1 , 7 , 28 ......

Here a = 0.6 d = 1.7 , 0.6 = 1.1, n = 24 

So , $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{24}=\frac{24}{2}[2 \times 0.6+(24-1)(1.1)]$

$=12[1.2+23 \times 1.1]$

$=12[1.2+25.3] 84$

$=12 \times 26.5$

$=318.0$

Question 2 

Ans: (i) $\begin{aligned}=& 34+32+30+\ldots+2 \\ & Here a=34, d=32-34=-2,1=2\\ & a_{n}=a+(n-1) d \\ \Rightarrow & 2=34+(n-1) \times(-2) \\ \Rightarrow & 2-34=-2(n-1) \\ \Rightarrow & \frac{-32}{-2}=n-1 \end{aligned}$

$\Rightarrow n-1=16$

$n=16+1$

$n=17$

So , $s_{n}=\frac{n}{2}(a+l) .$

$=\frac{17}{2}(34+2)$

$=\frac{17}{2} \times 36=306$

(ii) $7+9-\frac{1}{2}+12+\ldots+67$

Here, $a=7, d=9 \frac{1}{2}-7=2 \frac{1}{2}=\frac{5}{2}, 1=67$

$U=\left(a_{n}\right)=a+(n-1) d$

$\Rightarrow 67=7+(n-1)\left(\frac{5}{2}\right)$

$67-7=\frac{5}{2}(n-1)$

$\Rightarrow \frac{60 \times 2}{5}$

$=n-1$

$\Rightarrow n-1$

$=24$

= 24n = 24 +1 

=25 

So, 

$S_{25}=\frac{n}{2}[a+1]=\frac{25}{2}[7+67]$

=$\frac{25}{2} \times 74$

$=925$

Question 3

Ans: In an AP

$d=-2, a=100,1=-10$

$l=\left(a_{n}\right)=a+(n-1) d$

$-10=100+(n-1)(-2)$

$(n-1)(-2)=-10-100=-110$

$n-1=\frac{-110}{-2}=55$

$\Rightarrow n=55+1$

=56

$s_{56}=\frac{n}{2}[a+1]=\frac{56}{2}[100-10]$

$=28 \times 90$

$=2580$

Question 4

Ans: $\begin{aligned} \Rightarrow & \text { AP is } 54,51,48, \ldots \text { and } S_{n}=513 \\ & \text { Here, } a=54, d=51-54 \\=&-3 \end{aligned}$

$\begin{aligned} & S_{n}=\frac{n}{2}[2 a+(n-1) d] \\ & 513=\frac{n}{2}[2 \times 54+(n-1) \times(-3)] \\ \Rightarrow & 513 \times 2 \\=& n[108-3 n+3] \\ \Rightarrow & 10260 \\=& 108 n-3 n^{2}+3 n \\ \Rightarrow & 3 n^{2}-11 n+1026=0 \\ \Rightarrow & n^{2}-37 n+342=0 \\ \Rightarrow & n^{2}-18 n-19 n+342=0 \end{aligned}$

$\left\{\begin{array}{l}\because 342=-18 \times(-19) \\ -37=-18-19\end{array}\right\}$

$\begin{aligned} & \Rightarrow n(n-18)-19(n-18)=0 \\ \Rightarrow &(n-18)(n-19)=0 \end{aligned}$

Either n - 18 =0 , then n = 18

Or n- 19 =0 , then n = 19

So, Number of terms = 18 or 19 

Question 5

 

Ans: $\Rightarrow S_{9}=72, d=5$

let a be the first term,

n=9

So, $S_{g}=\frac{n}{2}[2 a+(n-1) d]$

$72=\frac{9}{2}[2 a+(9-1) \times 5]$

$\frac{72 \times 2}{9}=2 a+40$

$\Rightarrow 16=2 a+90$

$2 a=16-40=$

$2 a=-24$

$a=\frac{-24}{2}$

$a=-12$

 So a = -12

$a_{10}=a+(n-1) d$

$=-12+(10-1) \times 5$

$=-12+45$

$=-33$

Question 6

Ans: In an AP, 

Sum of first 6 terms 

=42

$a_{10}: a_{30}=1: 3$

Let a be the first term and d be the common difference , then

$S_{6}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 a+(6-1) d]$

$42=3(2 a+5 d)=6 a+15 d$

$\Rightarrow 6 a+15 d=42$

$a_{10}: a_{30}=1: 3$

$\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}$

$3 a+27 d=a+29 d$

$3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d$

$\Rightarrow a=d$

From (i) 

$6 a+15 a=42 \Rightarrow 21 a=42$

$\Rightarrow a=\frac{42}{21}=2$

$So, a=2, d=2$

So first term = 2

$a_{13}= a+(n-1) d=2+(13-1) \times 2$

$=2+12 \times 2=2+24$

$=26$

Question 7

Ans:  In a AP 

$a_{13}=4 \times a_{3}$

$a_{5}=16$

Let a be the first term and d be the common difference 

 So ,$\operatorname{a}_{5}=a+(n-1) d$

$=a+(5-1) d=a+4 d$

 So , $a+4 d=16$.............(i)

Similarly  

$a_{13}=a+12 d$ and $a_{3}=a+2 d$

$So, a+12 d=4 \times(a+2 d)$

$a+12 d=4 a+8 d$

$12 d-8 d=4 a-a \Rightarrow 3 a=4 d$

$a=\frac{4}{3} d$

From (i)

 $\frac{4}{3} d+4 d=16 \Rightarrow \frac{16}{3} d=16$

$\Rightarrow d=\frac{16 \times 3}{16}=3$

So, d= 3

and a = $\frac{4}{3} d=\frac{4}{3} \times 3=4$

$a=4, d=3$

Now $_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 4+(10-1) \times 3]$

$=5(8+27)$

$=5 \times 35=175$

Question 8

Ans:  $\Rightarrow A P$ is $8,10,12, \ldots$.

Were $a=8, d=10-8=2, n=60$

So  $T_{n}=a+(n-1) d$

$\Rightarrow 60=8+(n-1) \times 2$

$\Rightarrow(n-1) \times 2=60-8$

$=52$

$n-1=\frac{52}{2}$

$=26$

$n=26+1$

$=27$

$T_{60}=a+(60-1) d$

$=8+59 \times 2$

$=8+61=69$

Sum of last 10 term $=s_{60}-s_{50}$

$=\frac{60}{2}(2 a+59 d)-\frac{50}{2}(2 a+49 d)$

$=30(2 a+59 d)-25(2 a+49 d)$

$=60 a+1770 d-50 a-1225 d$

$=10 a+545 d=10 \times 8+545 \times 2$

$=80+1090$

$=1170$

Question 9

Ans: $\Rightarrow$ In an $A P$,

$\begin{aligned}&a_{12} 0 \times T_{12}=-13 \\&s_{4}=24\end{aligned}$

Let a be the first term and $b$ e the common difference, then

$a_{12}=a+(n-1) d$

$\Rightarrow-13=a+(12-1) d$

$\Rightarrow a+11 d=-13 \Rightarrow a=-13-11 d$

$s_{4}=\frac{n}{2}[2 a+(n-1) d]$

$24=\frac{4}{2}[2 a+3 d]=2[2 \times(-13-11 d)+3 d]$

$\frac{24}{2}=-26-22 d+3 d \Rightarrow 12=-26-19 d$

$\Rightarrow 12+26=-19 d \Rightarrow-19 d=38$

$d=\frac{38}{-19}=-2$

and $a=-13-11 d=-13+11 \times 2$

$=-13+22=9$

so, $a=9, d=-2$

Now , $s_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 9+(10-1)(-2)]$

$\begin{aligned} & 5[18+9(-2)]=5(18-18) \\=& 5 \times 0 . \\=& 0 \end{aligned}$

Question 10

Ans: $\because \Rightarrow$ Natural numbos betuteen 101 and 999 which are divisible by 2&5 both are 110 , $120,130, \ldots, 990$

Here $a,=110$ and $d=10,1=990$

 Now, $l=a_{n}=a+(n-1) d$ 

$990=110+(n-1) \times 10$

$\Rightarrow 990-110=10(n-1) \Rightarrow 10(n-1)=880$

$\Rightarrow n-1=\frac{880}{10} \Rightarrow n-1=88$

So , n = 88+ 1 = 89

Now , $s_{8 9}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{89}{2}[2 \times 110+(89-1) \times 10]$

$=\frac{89}{2}[220+88 \times 10]=\frac{89}{2}[220+880]$

$=\frac{89}{2} \times 110$

$=48950$

Question 11

Ans:  $\Rightarrow 2$ - digit number greater than 50 which are divisible by 7 , leaves a remainder of 4 are:

$53,60,67,74,81,88,95$

Here $a=53$,

$\begin{aligned} d &=7 \\ l &=95 \end{aligned}$

$\begin{aligned} \therefore a_{n}(l) &=a+(h-1) d . \\ 95 &=53+(h-1) \times 7 . \end{aligned}$

$\begin{aligned} 95-53 &=7(n-1) . \\ 42 &=7(n-1) \\ \frac{42}{7} &=(n-1) \\ 6 &=(n-1) \\(n-1) &=6 \\ n &=6+1 \\ n &=7 . \end{aligned}$

Then , 

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{7}{2}[2 \times 53+(7-1) \times 7]$

$=\frac{7}{2}[106+6 \times 7]$

$=\frac{7}{2}[106+42]$

$=\frac{7}{2} \times 48$

$=518$

Question 12

Ans: (i) Integers between 100 and 200 which are divisible by 9 are 108, 117 , 126 ,......198

Here 

a= 108

d =9

and l = 198

$\begin{aligned} \therefore a_{n}(l) &=a+(n-1) d . \\ 198 &=108+(n-1) \times 9 . \\ 198-108 &=9(n-1) \\ 90 &=9(n-1) \\ \frac{90}{9} &=n-1 \\ 10 &=n-1 \\ 10+1 &=n \\ n &=11 \\  & \end{aligned}$

Then 

$\begin{aligned} S_{11} &=\frac{h}{2}[2 a+(h-1) d] . \\ &=\frac{11}{2}[2 \times 108+(11-1) \times 9] \\ &=\frac{11}{2}[216+10 \times 9] \\ &=\frac{11}{2}[216+90] \end{aligned}$

$\frac{11}{2}\times 306$

=1683

(ii) Then sum of integers from 101 to 199

Here, a = 101,

d= 1 

l = 199

and n = 99

$s_{n}=$ $\frac{n}{2}[2 a+(n-1)d]$

$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$

$=\frac{99}{2}[202+98]$

$=\frac{99}{2}\times 300$

=14850

∴ Sum of integers which are not divisible by 9 

$=14850-1683$

$=13167$

Question 13

Ans:  Number between 9 and 95 when divided by 3, 

leaves remainder 1 

$10,13,16,19, \ldots . .99$

Here, 

$\begin{aligned} q &=10, \\ d &=3, \\ \text { and } l &=94 . \end{aligned}$

$a_{n}(l)=a+(n-1) d $

$94=10+(n-1) \times 3$

$94-10=3(n-1) .$

$84=3(n-1)$

$\frac{84}{3}=n-1$

$28=n-1$

$28+1=n$

$29=$n

$n=29$

Then, middle term 

$\begin{aligned} &=\frac{29+1}{2} \text { th } \\ &=\frac{30}{2} \mathrm{th} . \end{aligned}$

=15th term

Sum of first 14 terms 

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{74}{2}[2 \times 10+(14-1) \times 3]$

$=7[20+13 \times 3]$

$=7[20+39]$

$=7 \times 59$

$=413$

Middle terms = 10+15\times 3

= 10+45 = 55

After middle term; number of terms = 14 

Whose first terms (a)= 55 , d = 3

and n = 14

$\therefore s_{14}=\frac{19}{2}[2 \times 55+(14-1) \times 3]$

$\begin{aligned} &=7[110+39] \\ &=7 \times 149 \\ &=1043 \end{aligned}$

Question 14

Ans: $S_{n}=$ Sum of first n terms of an A.P.

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$To $ prove $=S_{12}=3\left(S_{8}-S_{4}\right)$

from $R \cdot H \cdot S=3\left[S_{8}-S_{4}\right]$

$=3\left[\frac{8}{2}[2 a+(8-1) d]-\frac{4}{2}[2 a+(4-1) d]\right] .$

$=3\left[\frac{8}{2}(2 a+7 d)-\frac{4}{2}[2 a+3 d)\right]$

$=3[4(2 a+7 d)-2(24+3 d)]$

$=3[84+28 d-4 a-6 d]$

$=3[4 a+22 d]$

$=12 a+66 d .$

$=6[2 a+11 d) .$

$\frac{12}{2}[2 a+(12-1) d]$= $S_{12}$ =L.H.S

Hence proved 

Question 15

Ans: Sum of first n even natural number $=\left(1+\frac{1}{h}\right)$ 

(Sum of first n odd natural numbers)

Sum of first n even natural number (2,4,6,8......2n)

$=\frac{n}{2}[2 a+(n-1) d] .$

$=\frac{n}{2}[2 \times 2+(n-1) \times 2]$

$=\frac{n}{2}[4+2n-2] .$

$=\frac{n}{2}[2+2n] .$

$=\frac{n}{2} \times 2(1+n)$

$=n(1+n) $

and Sum of n odd natural number (1, 3,5 ,.......(2n -1))

$=\frac{n}{2}[2 \times 1+(n-1) 2]$

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}[2 n]$

$=\frac{n}{2} \times 2[n]$

$=n^{2} .$

Then , 

$n^{2} \times \left(1+\frac{1}{n}\right)$

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

$=n^{2} \times \frac{n+1}{n}$

$=n(n+1)$

Question 16

Ans: Given,

First term $\left(a_{1}\right)=5$

Common difference $\left(d_{1}\right)=36$.

$S_{n}=$ Sum of $2_{n}$ terms of another $A P$ whose first term $\left(a_{2}\right)=36$ and common difference $\left(d_{2}\right)=5$.

 In first $A P$

$\begin{aligned} S_{n} &=\frac{n}{2}[2 a+(n-1) d] . \\ &=\frac{n}{2}[2 \times 5+(n-1) \times 36] \\ &=\frac{n}{2}[10+36 n-36] \\ &=\frac{n}{2}[36 n-26] . \end{aligned}$

$=\frac{n}{2} \times 2[18 n-13]$

$=n[18 n-13]$

In Second AP, 

$S_{2 N}=\frac{2 n}{2}[2 \times 36+(2 n-1) \times 5]$

$=n[72+10{n}-5]$

$=\mathrm{n}[67+10 \mathrm{n}] $

$\therefore S_{n}=S_{2 n}$

18 (18n- 13) = n(67 + 10n)

18 n - 13 =  $\frac{n}{n}(67+10 n)$

$18 n-10 n=67+13$

8n = 80

n = $\frac{80}{8}$

n=10

Question 17

Ans: Sum of first 10 terms of an AP = $4 \times sum$ of first 5 terms 

Let a be the first term and d be the common difference 

$\begin{aligned} \therefore S_{n} &=\frac{n}{2}[2 a+(n-1) d] \\ S_{10} &=\frac{10}{2}[2 a+(10-1) d] \\ &=5[2 a+9 d] \\ &=10a+45d \end{aligned}$

$\begin{aligned} S_{{5}} &=\frac{5}{2}[2 a+(5-1) d] \\ &=\frac{5}{2}[2 a+4 d] \end{aligned}$

$=\frac{5}{2} \times 2[a+2 d]$

$=5 a+10 d$

According to the question,

10a + 45 d =  $4 \times(5 a+10 d)$

$10 a+45 d=20 a+40 d .$

$45 d-40 b=20 a-10 a$

$5 a=109 .$

$\frac{5}{10}=\frac{a}{d}$

$\frac{a}{d}=\frac{5}{10}$

$\frac{a}{d}=\frac{1}{2}$

∴ Ratio in first term to the common difference = 1:2

Question 18

Ans:  Given, 

In the first row, plants are = 37

In second row = 35 

In third row = 33

and in the last row = 5

Here , 

a= 37 

d= 35 - 37 

d= - 2

$\begin{aligned} a_{n}(1)=& a+(n-1) d . \\ 5=& 37+(n-1) \times(-2) \\ 5 &-37=-2 n+2  \\ &-32=-2(n-1) . \\ & \frac{+32}{+2}=n-1 \end{aligned}$

$16=n-1$

$16+1=n$

$17=n$

$n=17$

So, number of rows = 17 

Then, 

Total plants $\left(S_{n}\right)=\frac{n}{2}[24+(n-1) d]$

$=\frac{17}{2}[2 \times 37+(17-1) \times(-2)] .$

$=\frac{17}{2}[74-32]$

=357

Hence , the total plants is 357

Question 19

Ans: Given , 

Total logs = 200 

In the bottom row, number of logs = 20 

and next row above it = 19 

Next row above it = 18

And Soon 

- AP is 20,19,18,17,16....... and $S_{n}=200$

Let the top row be the nth row 

$\therefore a_{n}=a+(n-1)d$

and $S_{n}=200$

So, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2 \times 20+(n-1) \times(-1)]$

$400=n[40-n+1]$

$400=40 n-n^{2}+h$

$400=41 n-n^{2} $

$n^{2}-41 n+400=0$

$n^{2}-25 n-16 n+400=0$

$n(n-25)-16(n-25)=0$

$(n-25)(n-16)=0$

n -25 =0 or n-16=0

n=25 or n =16

But n = 25 is not possible 

 As number of logs in the first row = 20 

and d = - 1 

Number of row = 16 

and number of log in 16th row 

$=9+(n-1) d$

$=20+(16-1) \times(-1)$

$=20-15$

$=5 logs$

Question 20

Ans: Giver,

Total amount= ₹1590,

Number of cash prizes = 7 

and each prize is Rs 50 less than the proceeding prize 

Let first prize = Rs a 

Then second prize = a - 50 

Third prize = a - 100

And so on 

$\therefore$ first term $=a$,

common difference (d) $=-50$,

$S_{n}=1890$

$n=7 $

$\begin{aligned} S_{n}=& \frac{n}{2}[2 a+(n-1) d] \\ 1890=& \frac{7}{2}[2 a+(7-1) \times(-50)] \\ 3780=& 7[2 a-300] \\ & \frac{3780}{7}=2 a-300 \\ 540=& 2 a-300 \\ 540+300=& 2 a \\ 840=& 2 a \\ \frac{840}{2}=9 \\ 420=a \\ a=420 \end{aligned}$

Hence prizes are $ ₹420, ₹ 370, ₹ 320 ; 270, ₹ 220, ₹ 170$ and $₹ 120$.