Class 11: Geometric Progression – Exercise 20.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Notes:

We know $S_n = a \Big($ $\frac{r^n - 1}{r-1}$ $\Big)$ when $r > 1$ and $S_n = a \Big($ $\frac{1- r^n }{1 - r}$ $\Big)$ when $r < 1$
$a_n = ar^{n-1}$

Question 1: Find the sum of the following geometric progressions:

i) $2, 6, 18, \ldots$ to $7$ terms

ii) $1, 3, 9, 27, \ldots$ to $8$ terms

iii) $1,$ $\frac{-1}{2}, \frac{1}{4}, \frac{-1}{8}$ $, \ldots$ to $9$ terms

iv) $( a^2 - b^2), (a-b),$ $\frac{a-b}{a+b}$ $, \ldots$ to $n$ terms

v) $4, 2, 1,$ $\frac{1}{2}$ $, \ldots$ to $10$ terms

Answer:

i) Given series: $2, 6, 18, \ldots$

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

$\therefore S_{7} = 2 \Big($ $\frac{3^7 - 1}{3 - 1}$ $\Big) = (3^7 -1) = 2186$

ii) Given series: $1, 3, 9, 27, \ldots$

Here $a = 1$ $r =$ $\frac{3}{1}$ $= 3$ $n = 8$

$\therefore S_{8} = 1 \Big($ $\frac{3^8 - 1}{3 - 1}$ $\Big) = 3280$

iii) Given series: $1,$ $\frac{-1}{2}, \frac{1}{4}, \frac{-1}{8}$ $, \ldots$

Here $a = 1$ $r =$ $\frac{-\frac{1}{2}}{1}$ $= \frac{-1}{2}$ $n = 9$

$\therefore S_{8} = 1 \Bigg($ $\frac{(\frac{-1}{2})^8 - 1}{(\frac{-1}{2}) - 1}$ $\Bigg) = \Bigg($ $\frac{\frac{-1}{512} -1}{\frac{-1}{2} -1}$ $\Bigg) =$ $\frac{171}{256}$

iv) Given series: $( a^2 - b^2), (a-b),$ $\frac{a-b}{a+b}$ $, \ldots$

Here $a = (a^2-b^2)$ $r =$ $\frac{a-b}{a^2-b^2} = \frac{a-b}{(a-b)(a+b)} = \frac{1}{a+b}$ $n = n$

$S_n = (a^2-b^2) \Big[$ $\frac{ ( \frac{1}{a+b} )^n - 1}{ \frac{1}{a+b}}$ $\Big]$

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

$=$ $\frac{(a-b)(a+b)[ 1 - ( a+b)^n ] (a+b) }{(a+b)^n [ 1 - (a + b)] }$

$=$ $\frac{(a-b)}{(a+b)^{n-2}} \cdot \Big[ \frac{1 - (a+b)^n }{1 - ( a+b) } \Big]$

$=$ $\frac{(a-b)}{(a+b)^{n-2}} \cdot \Big[ \frac{(a+b)^n - 1}{( a+b) - 1} \Big]$

v) Given series: $4, 2, 1,$ $\frac{1}{2}$ $, \ldots$

Here $a = 4$ $r =$ $\frac{2}{4}$ $=$ $\frac{1}{2}$ $n = 10$

$\therefore S_{10} = 4 \Bigg($ $\frac{1 - (\frac{1}{2})^{10}}{1- \frac{1}{2}}$ $\Bigg) = \Big(1 -$ $\frac{1}{1024}$ $\Big) =$ $\frac{1023}{128}$

$\\$

Question 2: Find the sum of the following geometric progressions:

i) $0.15 + 0.015 + 0.0015 + \ldots$ to $8$ terms

ii) $\sqrt{2} +$ $\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}$ $+ \ldots$ to $8$ terms

iii) $\frac{2}{9} - \frac{1}{3} + \frac{1}{2} - \frac{3}{4}$ $+ \ldots$ to $5$ terms

iv) $(x+y) + ( x^2 + xy + y^2) + ( x^3 + x^2y + xy^2 + y^3) + \ldots$ to $n$ terms

v) $\frac{3}{5} + \frac{4}{5^2} +\frac{3}{5^3} +\frac{4}{5^4}$ $+ \ldots$ to $2n$ terms

vi) $\frac{a}{(1+i)} + \frac{a}{(1+i)^2} + \frac{a}{(1+i)^3} + \ldots + \frac{a}{(1+i)^n}$

vii) $1, -a, a^2, - a^3 + \ldots$ to $n$ terms $(a \neq 1)$

viii) $x^3 , x^5, x^7 , \ldots$ to $n$ terms

ix) $\sqrt{7} + \sqrt{21} , 3\sqrt{7} + \ldots$ to $n$ terms

Answer:

i) Given sequence: $0.15 + 0.015 + 0.0015 + \ldots$

Here $a = 0.15$ $r =$ $\frac{0.015}{0.15}$ $= 0.1$ $n = 8$

$\therefore S_{8} = 0.15 \Big($ $\frac{1 - 0.1^8}{1- 0.1}$ $\Big) =$ $\frac{0.15}{0.9}$ $(1 - 0.1^8) =$ $\frac{1}{6}$ $(1-0.1^8) =$ $\frac{1}{6}$ $\Big( 1 -$ $\frac{1}{10^8}$ $\Big)$

ii) Given sequence: $\sqrt{2} +$ $\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}$ $+ \ldots$

Here $a = \sqrt{2}$ $r =$ $\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}$ $=$ $\frac{1}{2}$ $n = 8$

$\therefore S_8 = \sqrt{2} \Bigg($ $\frac{1 - (\frac{1}{2})^8}{1 - \frac{1}{2}}$ $\Bigg) = 2\sqrt{2} \Big( 1 -$ $\frac{1}{256}$ $\Big) =$ $\frac{255\sqrt{2}}{128}$

iii) Given sequence: $\frac{2}{9} - \frac{1}{3} + \frac{1}{2} - \frac{3}{4}$ $+ \ldots$

Here $a =$ $\frac{2}{9}$ $r =$ $\frac{\frac{-1}{3}}{\frac{2}{9}}$ $=$ $\frac{-3}{2}$ $n = 5$

$S_5 =$ $\frac{2}{9} \Big( \frac{1 - (\frac{-3}{2})^5}{1 - (\frac{-3}{2})} \Big) = \frac{2}{9} \Big( \frac{1 + \frac{3^5}{2^5} }{1 + \frac{3}{2} } \Big) = \frac{2}{9} \Big( \frac{(2^5+3^5) \cdot 2}{2^5 \cdot 5} \Big) = \frac{55}{72}$

iv) Given sequence: $(x+y) + ( x^2 + xy + y^2) + ( x^3 + x^2y + xy^2 + y^3) + \ldots$

$=$ $\frac{1}{(x-y)}$ $\Big[ (x^2 - y^2) + ( x^3 - y^3) + ( x^4 - y^4) + \ldots \Big]$

$\because \frac{x^n - y^n}{x-y }$ $= [ x^{n-1}+ x^{n-2} y + x^{n-3}y^{2} + \ldots + y^{n-1} ]$

$=$ $\frac{1}{(x-y)}$ $\Big[ (x^2 + x^3 + x^4 + \ldots) - ( y^2 + y^3 + y^4 + \ldots) \Big]$

$=$ $\frac{1}{(x-y)}$ $\Big[$ $\frac{x^2(x^n - 1)}{x-1}$ $-$ $\frac{y^2(y^n -1)}{y - 1}$ $\Big]$

v) Given sequence: $\frac{3}{5} + \frac{4}{5^2} +\frac{3}{5^3} +\frac{4}{5^4}$ $+ \ldots$

$= \frac{3}{5} \Big[ \Big(1 + \frac{1}{5^2} + \frac{1}{5^4} + \ldots \Big) + \frac{4}{5^2} \Big( 1 + \frac{1}{5^2} + \frac{1}{5^4} + \ldots \Big) \Big]$

$=$ $\Big( \frac{3}{5} + \frac{4}{5^2} \Big) \Big( 1 + \frac{1}{5^2} + \frac{1}{5^4} + \ldots\Big)$

$=$ $\frac{19}{25} \Big[ \frac{1 - ( \frac{1}{5^2})^n}{1 - \frac{1}{5^2}} \Big]$

$=$ $\frac{19}{25} \Big[ 1 - \frac{1}{5^{2n}} \Big]$

vi) Given sequence: $\frac{a}{(1+i)} + \frac{a}{(1+i)^2} + \frac{a}{(1+i)^3} + \ldots + \frac{a}{(1+i)^n}$

Here $a =$ $\frac{a}{1+i}$ $r =$ $\frac{\frac{a}{(1+i)^2}}{\frac{a}{(i)}}$ $=$ $\frac{1}{1+i}$ $n = n$

$S_n =$ $\frac{a}{1+i}$ $\Bigg[$ $\frac{1 - (\frac{1}{1+i})^n }{1 - (\frac{1}{1+i})}$ $\Bigg]$

$=$ $\frac{a}{1+i}$ $\Bigg[$ $\frac{[(1+i)^n-1](i+i) }{(1+i)^n(1+i-1)}$ $\Bigg]$

$=$ $\frac{a[ (1+i)^n - 1}{i(1+i)^n}$

$=$ $\frac{ai}{-1}$ $[ 1 - ( 1+i)^{-n} ]$

$= -ai [ 1 - ( 1+i)^{-n} ]$

vii) Given sequence: $1, -a, a^2, - a^3 + \ldots$

Here $a = 1$ $r =$ $\frac{-a}{1}$ $= -a$ $n = n$

$S_n = 1 \Big[$ $\frac{1 - ( -a)^n}{1 - ( -a)}$ $\Big] =$ $\frac{1 - ( -a)^n}{1+a}$

viii) Given sequence: $x^3 , x^5, x^7 , \ldots$

Here $a = x^3$ $r =$ $\frac{x^5}{x^3}$ $= x^2$ $n = n$

$S_n = x^3 \Big[$ $\frac{1 - ( x^2)^n}{1 - ( x^2)}$ $\Big] =$ $\frac{x^3[(1 - x^{2n} )}{1-x^2}$ $\Big] =$ $\frac{x^3[(x^{2n}-1 )}{x^2-1}$

ix) Given sequence: $\sqrt{7} + \sqrt{21} , 3\sqrt{7} + \ldots$

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

$S_n = \sqrt{7} \Big[$ $\frac{( \sqrt{7} )^n-1}{\sqrt{7}-1}$ $\Big]$

$\\$

Question 3: Evaluate the following:

i) $\sum \limits_{n=1}^{11} (2+3^n)$ ii) $\sum \limits_{k=1}^{n} (2^k+3^{k-1})$ iii) $\sum \limits_{n=2}^{10} 4^n$

Answer:

i) $\sum \limits_{n=1}^{11} (2+3^n)$

$= ( 2 + 3^{1}) + ( 2 + 3^{2}) + ( 2 + 3^{3}) + \ldots+ ( 2 + 3^{11})$

$= 2 \times 11 + ( 3^{1}+ 3^{2}+ 3^{3}+ \ldots + 3^{11})$

$= 22 +$ $\frac{3}{2}$ $\Big($ $\frac{3^{11}-1}{3-1}$ $\Big)$

$= 22 +$ $\frac{3}{2}$ $(3^{11} - 1)$

$= 265741$

ii) $\sum \limits_{k=1}^{n} (2^k+3^{k-1})$

$= ( 2^{1}+3^{0}) + ( 2^{2}+3^{1}) + ( 2^{3}+3^{2}) + \ldots + ( 2^{n}+3^{n-1})$

$= ( 2^{1} + 2^{2} + 2^{3} + \ldots + 2^{n} ) + ( 3^{0}+3^{1}+3^{2}+ \ldots + 3^{n-1})$

$= 2 \Big($ $\frac{2^n - 1}{2 - 1}$ $\Big) + 1 \Big($ $\frac{3^n - 1}{3 - 1}$ $\Big)$

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

$=$ $\frac{1}{2}$ $( 2^{n+2} + 3^n - 5)$

iii) $\sum \limits_{n=2}^{10} 4^n$

$= 4^2 + 4^3 + \ldots + 4^{10}$

$= 4^2 ( 1 + 4 + 4^2 + \ldots + 4^8)$

$= 4^2 \Big($ $\frac{4^9 - 1}{4-1}$ $\Big)$

$=$ $\frac{16}{3}$ $( 4^9 - 1)$

$= 1398096$

$\\$

Question 4: Find the sum of the following series:

i) $5 + 55 + 555 + \ldots$ to $n$ terms

ii) $7 + 77 + 777 + \ldots$ to $n$ terms

iii) $9 + 99 + 999 + \ldots$ to $n$ terms

iv) $0.5+0.55+0.555+ \ldots$ to $n$ terms

v) $0.6 + 0.66 + 0.666 + \ldots$ to $n$ terms

Answer:

i) $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms

$= 5 [ 1 + 1 + 111 + \ldots \text{ n terms } ]$

$=$ $\frac{5}{9}$ $[ 9 + 99 + 999 + \ldots \text{ n terms } ]$

$=$ $\frac{5}{9}$ $[ ( 10 - 1) + ( 10^2-1) + ( 10^3 - 1) + \ldots + ( 10^n -1) ]$

$=$ $\frac{5}{9}$ $[ ( 10 + 10^2 + 10^3 + \ldots + 10^n) - n ]$

$=$ $\frac{5}{9}$ $\Big[$ $\frac{10(10^n-1)}{10-1}$ $- n \Big]$

$=$ $\frac{5}{9}$ $\Big[$ $\frac{10}{9}$ $(10^n - 1) - n \Big]$

$=$ $\frac{5}{81}$ $[ 10^{n+1}-9n -10]$

ii) $S_n = 7 + 77 + 777 + \ldots$ to $n$ terms

$= 7 [ 1 + 1 + 111 + \ldots \text{ n terms } ]$

$=$ $\frac{7}{9}$ $[ 9 + 99 + 999 + \ldots \text{ n terms } ]$

$=$ $\frac{7}{9}$ $[ ( 10 - 1) + ( 10^2-1) + ( 10^3 - 1) + \ldots + ( 10^n -1) ]$

$=$ $\frac{7}{9}$ $[ ( 10 + 10^2 + 10^3 + \ldots + 10^n) - n ]$

$=$ $\frac{7}{9}$ $\Big[$ $\frac{10(10^n-1)}{10-1}$ $- n \Big]$

$=$ $\frac{7}{9}$ $\Big[$ $\frac{10}{9}$ $(10^n - 1) - n \Big]$

$=$ $\frac{7}{81}$ $[ 10^{n+1}-9n -10]$

iii) $S_n = 9 + 99 + 999 + \ldots$ to $n$ terms

$= 9 [ 1 + 1 + 111 + \ldots \text{ n terms } ]$

$=$ $[ 9 + 99 + 999 + \ldots \text{ n terms } ]$

$=$ $[ ( 10 - 1) + ( 10^2-1) + ( 10^3 - 1) + \ldots + ( 10^n -1) ]$

$=$ $[ ( 10 + 10^2 + 10^3 + \ldots + 10^n) - n ]$

$=$ $\Big[$ $\frac{10(10^n-1)}{10-1}$ $- n \Big]$

$=$ $\Big[$ $\frac{10}{9}$ $(10^n - 1) - n \Big]$

$=$ $\frac{1}{9}$ $[ 10^{n+1}-9n -10]$

iv) $S_n = 0.5+0.55+0.555+ \ldots$ to $n$ terms

$= 5 [ 0.1+0.11+0.111+ \ldots \text{ n terms } ]$

$=$ $\frac{5}{9}$ $\Big[$ $\frac{9}{10}$ $+$ $\frac{9}{100}$ $+$ $\frac{9}{1000}$ $+ \ldots \text{ n terms } \Big]$

$=$ $\frac{5}{9}$ $\Big[ \Big( 1-$ $\frac{1}{10}$ $\Big) +\Big( 1-$ $\frac{1}{10^2}$ $\Big) +\Big( 1-$ $\frac{1}{10^3}$ $\Big) + \ldots \text{ n terms } ]$

$=$ $\frac{5}{9}$ $\Big[ n -$ $\frac{1}{10}$ $\Big($ $\frac{1}{10}$ $+$ $\frac{1}{10^2}$ $+$ $\frac{1}{10^3}$ $+\ldots \text{ n terms } \Big]$

$=$ $\frac{5}{9}$ $\Big[ n -$ $\frac{1}{10} \Big( \frac{1 - (\frac{1}{10^n} ) }{1 - \frac{1}{10} }\Big)$ $\Big]$

$=$ $\frac{5}{9}$ $\Big[ n -$ $\frac{1}{10} \Big( \frac{(10^n - 1) 10}{10^n ( 10-1)} \Big)$ $\Big]$

$=$ $\frac{5}{9}$ $\Big[ n -$ $\frac{1}{9} \Big( \frac{(10^n - 1) }{10^n } \Big)$ $\Big]$

$=$ $\frac{5}{9}$ $\Big[ n -$ $\frac{1}{9} \Big( 1 - \frac{1 }{10^n } \Big)$ $\Big]$

v) $S_n = 0.6 + 0.66 + 0.666 + \ldots$ to $n$ terms

$= 6 [ 0.1+0.11+0.111+ \ldots \text{ n terms } ]$

$=$ $\frac{6}{9}$ $\Big[$ $\frac{9}{10}$ $+$ $\frac{9}{100}$ $+$ $\frac{9}{1000}$ $+ \ldots \text{ n terms } \Big]$

$=$ $\frac{6}{9}$ $\Big[ \Big( 1-$ $\frac{1}{10}$ $\Big) +\Big( 1-$ $\frac{1}{10^2}$ $\Big) +\Big( 1-$ $\frac{1}{10^3}$ $\Big) + \ldots \text{ n terms } ]$

$=$ $\frac{6}{9}$ $\Big[ n -$ $\frac{1}{10}$ $\Big($ $\frac{1}{10}$ $+$ $\frac{1}{10^2}$ $+$ $\frac{1}{10^3}$ $+\ldots \text{ n terms } \Big]$

$=$ $\frac{6}{9}$ $\Big[ n -$ $\frac{1}{10} \Big( \frac{1 - (\frac{1}{10^n} ) }{1 - \frac{1}{10} }\Big)$ $\Big]$

$=$ $\frac{6}{9}$ $\Big[ n -$ $\frac{1}{10} \Big( \frac{(10^n - 1) 10}{10^n ( 10-1)} \Big)$ $\Big]$

$=$ $\frac{6}{9}$ $\Big[ n -$ $\frac{1}{9} \Big( \frac{(10^n - 1) }{10^n } \Big)$ $\Big]$

$=$ $\frac{6}{9}$ $\Big[ n -$ $\frac{1}{9} \Big( 1 - \frac{1 }{10^n } \Big)$ $\Big]$

$\\$

Question 5: How many terms of the G.P. $3, 3/2, 3/4, \ldots$ to be taken together to make $\frac{3069}{512}$ ?

Answer:

Here we have $a = 3$ $r =$ $\frac{\frac{3}{2}}{3}$ $=$ $\frac{1}{2}$ $S_n =$ $\frac{3069}{512}$

$\frac{3069}{512}$ $= 3 \Big($ $\frac{1 - (\frac{1}{2})^n}{1 - \frac{1}{2}}$ $\Big)$

$\Rightarrow$ $\frac{3069}{512}$ $= 6 \Big( 1 - \Big($ $\frac{1}{2}$ $\Big)^n \Big)$

$\Rightarrow$ $\frac{3069}{3072}$ $= 1 - \Big($ $\frac{1}{2}$ $\Big)^n$

$\Rightarrow \Big($ $\frac{1}{2}$ $\Big)^n= 1 -$ $\frac{3069}{3072}$ $=$ $\frac{3}{3072}$ $=$ $\frac{1}{1024}$ $=$ $\frac{1}{2^{10}}$

$\Rightarrow n = 10$

$\\$

Question 6: How many terms of the series $2 + 6 + 18 + \ldots$ must be taken to make the sum equal to $728$ ?

Answer:

Here we have $a = 2$ $r =$ $\frac{6}{2}$ $= 3$ $S_n = 728$

$728 = 2 ($ $\frac{3^n - 1}{3-1}$ $)$

$\Rightarrow 3^n = 729$

$\Rightarrow 3^n = 3^6$

$\Rightarrow n = 6$

$\\$

Question 7: How many terms of the sequence $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ must be taken to make the sum $39 + 13\sqrt{3}$ ?

Answer:

Here we have $a = \sqrt{3}$ $r =$ $\frac{3}{\sqrt{3}}$ $= \sqrt{3}$ $S_n = 39 + 13\sqrt{3}$

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

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

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

$\Rightarrow (\sqrt{3})^n - 1 =$ $\frac{26\sqrt{3}}{\sqrt{3}}$ $= 26$

$\Rightarrow (\sqrt{3})^n = 26 + 1 = 27 = (\sqrt{3})^6$

$n = 6$

$\\$

Question 8: The sum of $n$ terms of the G.P. is $3 , 6, 12 \ldots$ is 381. Find the value of $n$ .

Answer:

Here we have $a = 3$ $r =$ $\frac{6}{3}$ $= 2$ $S_n = 381$

$381 = 3 \Big($ $\frac{2^n - 1}{2-1}$ $\Big)$

$\Rightarrow 2^n = 127 + 1 = 128 = 2^7$

$\Rightarrow n = 7$

$\\$

Question 9: The common ratio of a G.P. is $3$ and the last term is $486$ . If the sum of these terms be $728$ , find the first term.

Answer:

Here we have $r = 3$ $a_n = 486$ $S_n = 728$

We know $a_n = ar^{n-1}$

$\Rightarrow 486 = a \cdot 3^{n-1}$ … … … … … i)

Also, $728 = a ($ $\frac{3^n - 1}{3 -1}$ $)$

$\Rightarrow 1456 = a ( 3^n -1)$ … … … … … ii)

Dividing ii) by i) we get

$\frac{1456}{486}$ $=$ $\frac{3^n - 1}{3^{n-1}}$

$\Rightarrow$ $\frac{1456}{486}$ $= 3 -$ $\frac{1}{3^{n-1}}$

$\Rightarrow$ $\frac{1}{3^{n-1}}$ $= 3 -$ $\frac{1456}{486}$ $=$ $\frac{2}{486}$ $=$ $\frac{1}{243}$ $=$ $\frac{1}{3^5}$

$\Rightarrow 3^{n-1} = 3^5$

$\Rightarrow n = 6$

$\\$

Question 10: The ratio of the sum of first three terms is to that of first six terms of a G.P. is $125:152$ . Find the common ratio.

Answer:

Given $\frac{S_3}{S_6} = \frac{125}{152}$

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

$\therefore \frac{a(\frac{r^3-1}{r-1})}{a( \frac{r^6-1}{r-1} )} = \frac{125}{152}$

$\frac{r^3-1}{r^6-1} = \frac{125}{152}$

$125 ( r^6-1) = 152 ( r^3 - 1)$

$125r^6 - 152 r^3 +27 = 0$

Let $r^3 = x$

$\therefore 125 x^2 - 152 x + 27 = 0$

$\Rightarrow ( x - 1) ( 125x - 27) = 0$

$x = 1$ or $x =$ $\frac{27}{125}$

$\therefore r^3 = 1 \Rightarrow r = 1$ . But $r \neq 1$

Hence $r^3 =$ $\frac{27}{125}$ $= \Big($ $\frac{3}{5}$ $\Big)^3 \Rightarrow r =$ $\frac{3}{5}$

$\\$

Question 11: The $4^{th}$ and the $7^{th}$ term of a G.P. are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of n terms of the G.P.

Answer:

Given $a_4 =$ $\frac{1}{27}$ and $a_7 =$ $\frac{1}{729}$

$\therefore ar^3 =$ $\frac{1}{27}$ … … … … … i)

and $ar^6 =$ $\frac{1}{729}$ … … … … … ii)

Dividing ii) by i) we get

$\frac{ar^6}{ar^3}$ $=$ $\frac{27}{729}$ $=$ $\frac{1}{27}$

$r^3 = \Big($ $\frac{1}{3}$ $\Big )^3$

$\therefore r =$ $\frac{1}{3}$

Substituting in i) we get $a = 1$

$\therefore S_n = 1 \Big($ $\frac{1 - ( \frac{1}{3})^n}{1-\frac{1}{3}}$ $\Big ) =$ $\frac{3}{2}$ $\Big ( 1 -$ $\frac{1}{3^n}$ $\Big)$

$\\$

Question 12: Find the sum $\sum \limits_{n=1}^{10} \Big \{ \Big( \frac{1}{2} \Big)^{n-1} + \Big( \frac{1}{5} \Big)^{n+1} \Big \}$

Answer:

$\sum \limits_{n=1}^{10} \Big \{ \Big( \frac{1}{2} \Big)^{n-1} + \Big( \frac{1}{5} \Big)^{n+1} \Big \}$

$= \Big[ \Big( \frac{1}{2} \Big)^0 + \Big( \frac{1}{5} \Big)^2 \Big] + \Big[ \Big( \frac{1}{2} \Big)^1 + \Big( \frac{1}{5} \Big)^3 \Big] +\ldots + \Big[ \Big( \frac{1}{2} \Big)^9 + \Big( \frac{1}{5} \Big)^{11} \Big]$

$= \Big[ \frac{1}{2^0} + \frac{1}{2^1} + \ldots + \frac{1}{2^9} \Big] + \Big[ \frac{1}{5^2} + \frac{1}{5^3} + \ldots + \frac{1}{5^{11}} \Big]$

$= 1\Bigg[$ $\frac{1 - (\frac{1}{2})^{10}}{1 - \frac{1}{2}}$ $\Bigg] +$ $\frac{1}{5^2}$ $\Bigg[$ $\frac{1 - (\frac{1}{5})^{10}}{1 - \frac{1}{5}}$ $\Bigg]$

$= 2 \Big(1 -$ $\frac{1}{2^{10}}$ $\Big) +$ $\frac{1}{4 \cdot 5}$ $\Big( 1 -$ $\frac{1}{5^{10}}$ $\Big)$

$=$ $\frac{2^{10}-1}{2^9}$ $+$ $\frac{5^{10}-1}{4 \cdot 5^{11}}$

$\\$

Question 13: The fifth term of a G.P. is $81$ whereas its second term is $24$ . Find the series and sum of its first eight terms.

Answer:

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

Given $a_2 = 24$ $\Rightarrow ar= 24$ … … … … … i)

Similarly, $a_5 = 81$ $\Rightarrow ar^4 = 81$ … … … … … ii)

Dividing ii) by i) we get

$\frac{ar^4}{ar}$ $=$ $\frac{81}{24}$

$\Rightarrow r^3 =$ $\frac{27}{8}$ $= \Big($ $\frac{3}{2}$ $\Big)^3$

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

Substituting in i) we get $a =$ $\frac{24}{3/2}$ $= 16$

$\therefore S_8 = 16 \Bigg($ $\frac{(\frac{3}{2})^8 - 1}{\frac{3}{2} -1 }$ $\Bigg) = 32 \Big($ $\frac{3^8}{2^8}$ $- 1 \Big) = 32 \Big($ $\frac{6561-256}{256}$ $\Big ) =$ $\frac{6305}{8}$

$\\$

Question 14: If $S_1, S_2, S_3$ be respectively the sums of $n, 2n , 3n$ terms of a G.P., then prove that ${S_1}^2+ {S_2}^2 = S_1 ( S_2 + S_3)$ .

Answer:

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

Given $S_1 = a \Big($ $\frac{r^n - 1}{r - 1}$ $\Big )$

$S_2 = a \Big($ $\frac{r^{2n} - 1}{r - 1}$ $\Big ) =$ $\frac{a (r^n - 1)( r^n + 1)}{r-1}$ $= S_1 ( r^n +1)$

$S_3 = a \Big($ $\frac{r^{3n} - 1}{r - 1}$ $\Big ) =$ $\frac{a (r^n - 1)( r^{2n} + r^n + 1)}{r-1}$ $= S_1 ( r^{2n} + r^n + 1)$

To prove: ${S_1}^2+ {S_2}^2 = S_1 ( S_2 + S_3)$

LHS $= {S_1}^2+ {S_2}^2$

$= {S_1}^2+ {S_1}^2 ( r^{n} + 1)^2$

$= {S_1}^2 ( 1 + r^{2n} + 2r^n + 1)$

$= S_1 [ S_1 ( r^{2n} + r^n + 1) + S_1 ( 1 + r^n)]$

$= S_1 [ S_2 + S_3] =$ RHS. Hence proved.

$\\$

Question 15: Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $\frac{1}{r^n}$ .

Answer:

For the first $n$ terms

Let the first term be $a$ and the common ratio is $r$ .

$\therefore S_n = a \Big($ $\frac{r^n - 1}{r-1}$ $\Big )$ … … … … … i)

For next $n$ terms

First term will be $a_{n+1}$ th and the common ratio is $r$ .

$S_{next \ nth \ terms} = ar^n \Big($ $\frac{r^n - 1}{r-1}$ $\Big )$ … … … … … ii)

$\therefore$ $\frac{S_n}{S_{ next \ nth \ terms}} = \frac{a( \frac{r^n - 1}{r-1} )}{ar^n ( \frac{r^n - 1}{r-1} )} = \frac{1}{r^n}$

Hence proved.

$\\$

Question 16: If $a , b$ are the roots of $x^2 - 3x + p = 0$ and $c, d$ are the roots of $x^2 - 12 x + q = 0$ , where $a, b, c , d$ form a G.P. Prove that $(q+p):(q-p) = 17:15$

Answer:

Given $a , b$ are the roots of $x^2 - 3x + p = 0$

$\Rightarrow a+b = - 3$ … … … … … i) and $ab = p$ … … … … … ii)

Similarly, $c, d$ are the roots of $x^2 - 12 x + q = 0$

$\Rightarrow c+d = 12$ … … … … … iii) and $cd = q$ … … … … … iv)

Now $a, b, c, d$ are in G.P.

Let the common ratio be $r$ . Therefore

$b = ar, \hspace{1.0cm} c = ar^2, \hspace{1.0cm} d = ar^3$

Substituting in i) & iii) we get

$a + ar = 3 \Rightarrow a ( 1 + r) = 3$ … … … … … v)

$ar^2 + ar^3 = 12 \Rightarrow ar^2( 1+r) = 12$ … … … … … vi)

Dividing vi) by v) we get

$\frac{ar^2( 1+r)}{a ( 1 + r)}$ $=$ $\frac{12}{3}$

$\Rightarrow r^2 = 4 \Rightarrow r = 2$

$\therefore$ v) , $a = \frac{3}{1+2} = 1$

$\therefore p = ab = (1)(1 \cdot 2^1) = 2$ and $q = cd = (1 \cdot 2^2) (1 \cdot 2^3) = 2^5 = 32$

Hence $\frac{q+p}{q-p} =\frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15}$

$\\$

Question 17: How many terms of the G.P. $3,$ $\frac{3}{2}, \frac{3}{4}$ $, \ldots$ are needed to give the sum $\frac{3069}{512}$ ?

Answer:

We have $a = 3$ $r =$ $\frac{3/2}{3}$ $=$ $\frac{1}{2}$ $S_n =$ $\frac{3069}{512}$

$\therefore$ $\frac{3069}{512}$ $= 3 \Big($ $\frac{1 - (\frac{1}{2})^n}{1 -\frac{1}{2} }$ $\Big)$

$\Rightarrow$ $\frac{3069}{512}$ $= 6 ( 1 -$ $\frac{1}{2^n}$ $)$

$\Rightarrow$ $\frac{3069}{3072}$ $= 1 -$ $\frac{1}{2^n}$

$\Rightarrow$ $\frac{1}{2^n}$ $= 1 -$ $\frac{3069}{3072}$ $=$ $\frac{1}{1024}$ $=$ $\frac{1}{2^{10}}$

$\Rightarrow n =10$

$\\$

Question 18: A person has $2$ parents, $4$ grandparents, $8$ great grand parents, and so on. Find the number of his ancestors during the $10$ generations proceeding his own.

Answer:

We have $a = 3$ $r =$ $\frac{2^2}{2}$ $= 2$

$\therefore S_{10} = 2 \Big($ $\frac{2^{10}-1}{2-1}$ $\Big) = 2(2^{10}-1) = 2 \times 1023 = 2046$

$\\$

Question 19: If $S_1, S_2, \ldots , S_n$ are the sums of $n$ terms of $n$ G.P.s whose first term is $1$ in each and common ratios are $1, 2, 3, \ldots, n$ respectively, then prove that $S_1+S_2+2S_3 +3S_4 + \ldots (n-1) S_n = 1^n + 2^n + 3^n + \ldots + n^n$

Answer:

Given $S_1, S_2, \ldots , S_n$ are the sums of $n$ terms of $n$ G.P.s whose first term is $1$ in each and common ratios are $1, 2, 3, \ldots, n$ respectively

For $S_1, a = 1$ and $r = 1 \therefore S_1 = 1 + 1 + 1 + \ldots + 1 = n$

Similarly, $S_2 = 1 \Big($ $\frac{2^n - 1}{2-1}$ $\Big) =$ $\frac{(2^n - 1)}{1}$

$S_3 = 1 \Big($ $\frac{3^n - 1}{3-1}$ $\Big) =$ $\frac{(3^n - 1)}{2}$

$S_4 = 1 \Big($ $\frac{4^n - 1}{4-1}$ $\Big) =$ $\frac{(4^n - 1)}{3}$

$\ldots$

$S_n = 1 \Big($ $\frac{n^n - 1}{n-1}$ $\Big) =$ $\frac{(n^n - 1)}{n-1}$

Now $S_1+S_2+2S_3 +3S_4 + \ldots (n-1) S_n$

$= n + 1 \cdot$ $\frac{(2^n - 1)}{1}$ $+ 2 \cdot$ $\frac{(3^n - 1)}{2}$ $+ 3 \cdot$ $\frac{(4^n - 1)}{3}$ $+ \ldots + (n-1) \cdot$ $\frac{(n^n - 1)}{n-1}$

$= n + ( 2^n - 1) + ( 3^n - 1) + ( 4^n - 1) + \ldots + ( n^n - 1)$

$= n + ( 2^n + 3^n + 4^n + \ldots + n^n) - ( n - 1)$

$= 1 + 2^n + 3^n + 4^n + \ldots + n^n =$ RHS. Hence proved.

$\\$

Question 20: A G.P. consists of even number of terms. If the sum of all the terms is $5$ times the sum of the terms occupying the odd places. Find the common ratio of the G.P.

Answer:

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

Let there be $2n$ terms in the G.P.

Sum of all the terms $= 5$ ( Sum of the terms occupying the odd places)

$\Rightarrow a_1 + a_2 + a_ 3 + \ldots + a_{2n} = 5 ( a _ 1 + a_3 + a_5 + \ldots + a_{2n-1})$

$\Rightarrow a + ar + ar^2 + \ldots + ar^{2n-1} = 5a \Big($ $\frac{1 - (r^2)^n}{1-r^2}$ $\Big)$

$\Rightarrow a \Big($ $\frac{1 - r^{2n}}{1-r}$ $\Big) = 5a \Big($ $\frac{1 - r^{2n}}{1-r}$ $\Big) \Big($ $\frac{1}{1+r}$ $\Big)$

$\Rightarrow 1 + r = 5$

$\Rightarrow r = 4$

$\\$

Question 21: Let $a_n$ be the $n^{th}$ term of the G.P. of positive numbers. Let $\sum \limits_{n=1}^{100} a_{2n} = \alpha$ and $\sum \limits_{n=1}^{100} a_{2n-1} = \beta$ , such that $\alpha \neq \beta$ . Prove that the common ratio of the G.P. is $\frac{\alpha }{ \beta}$ .

Answer:

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

$\sum \limits_{n=1}^{100} a_{2n} = \alpha$

$\therefore a_2 + a_4 + \ldots + a_{200} = \alpha$

$\Rightarrow ar + ar^3 + \ldots + ar^{199} = \alpha$

$\Rightarrow ar \Big($ $\frac{1 - ( r^2)^{100}}{1 - r^2}$ $\Big) = \alpha$ … … … … … i)

$\sum \limits_{n=1}^{100} a_{2n-1} = \beta$

$\therefore a_1 + a_3 + \ldots + a_{199} = \beta$

$\Rightarrow a + ar^2 + \ldots + ar^{198} = \beta$

$\Rightarrow a \Big($ $\frac{1 - ( r^2)^{100}}{1 - r^2}$ $\Big) = \beta$ … … … … … ii)

$\frac{\alpha}{\beta} = \frac{ ar \Big(\frac{1 - ( r^2)^{100}}{1 - r^2} \Big)}{a \Big( \frac{1 - ( r^2)^{100}}{1 - r^2} \Big) }$ $= r$

$\therefore r =$ $\frac{\alpha}{\beta}$

$\\$

Question 22: Find the sum of $2n$ terms of the series whose every even term is $'a'$ times the term before it and every odd term is $'c'$ times the term before it, the first term being unity.

Answer:

Let the given series be $a_1+a_2+a_3 +a_4 + \ldots + a_n$

Given $a_1 = 1, \hspace{0.5cm} a_2 = a \cdot a_1, \hspace{0.5cm} a_3 = c \cdot a_2, \hspace{0.5cm} a_4 = a \cdot a_3, \hspace{0.5cm} a_5 = c \cdot a_4 \ldots$

$\therefore S_n = a_1+a_2+a_3 +a_4 + \ldots + a_n$

$= 1+a + ac + a^2c + a^2c^2 + \ldots \text{2n terms}$

$= (1+a) + ac( 1+a) + a^2c^2 ( 1+a) + \ldots \text{2n terms}$

$= (1+a) \Big($ $\frac{1-(ac)^n}{1-ac}$ $\Big)$

$= (1+a) \Big($ $\frac{(ac)^n-1}{ac-1}$ $\Big)$

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Class 11: Geometric Progression – Exercise 20.3

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