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

Notes:

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

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

$\displaystyle \text{i) } 2, 6, 18, \ldots \text{ to } 7 \text{ terms }$

$\displaystyle \text{ii) } 1, 3, 9, 27, \ldots \text{ to } 8 \text{ terms }$

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

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

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

Answer:

$\displaystyle \text{i) } \text{Given series: } 2, 6, 18, \ldots$

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

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

$\displaystyle \text{ii) } \text{Given series: } 1, 3, 9, 27, \ldots$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{i) } 0.15 + 0.015 + 0.0015 + \ldots \text{ to } 8 \text{ terms }$

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

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

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

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

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

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

$\displaystyle \text{viii) } x^3 , x^5, x^7 , \ldots \text{ to } n \text{ terms }$

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

Answer:

$\displaystyle \text{i) } \text{Given sequence: } 0.15 + 0.015 + 0.0015 + \ldots$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \text{viii) } \text{Given sequence: } x^3 , x^5, x^7 , \ldots$

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

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

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

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

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

$\displaystyle \\$

Question 3: Evaluate the following:

$\displaystyle \text{i) } \sum \limits_{n=1}^{11} (2+3^n) \hspace{1.0cm} \text{ii) } \sum \limits_{k=1}^{n} (2^k+3^{k-1}) \hspace{1.0cm} \text{iii) } \sum \limits_{n=2}^{10} 4^n$

Answer:

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

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

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

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

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

$\displaystyle = 265741$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 1398096$

$\displaystyle \\$

Question 4: Find the sum of the following series:

$\displaystyle \text{i) } 5 + 55 + 555 + \ldots \text{ to } n \text{ terms }$

$\displaystyle \text{ii) } 7 + 77 + 777 + \ldots \text{ to } n \text{ terms }$

$\displaystyle \text{iii) } 9 + 99 + 999 + \ldots \text{ to } n \text{ terms }$

$\displaystyle \text{iv) } 0.5+0.55+0.555+ \ldots \text{ to } n \text{ terms }$

$\displaystyle \text{v) } 0.6 + 0.66 + 0.666 + \ldots \text{ to } n \text{ terms }$

Answer:

$\displaystyle \text{i) } S_n = 5 + 55 + 555 + \ldots \text{ to } n \text{ terms }$

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

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

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

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

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

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

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

$\displaystyle \text{ii) } S_n = 7 + 77 + 777 + \ldots \text{ to } n \text{ terms }$

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

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

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

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

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

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

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

$\displaystyle \text{iii) } S_n = 9 + 99 + 999 + \ldots \text{ to } n \text{ terms }$

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

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

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

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

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

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

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

$\displaystyle \text{iv) } S_n = 0.5+0.55+0.555+ \ldots \text{ to } n \text{ terms }$

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

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

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

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

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

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

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

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

$\displaystyle \text{v) } S_n = 0.6 + 0.66 + 0.666 + \ldots \text{ to } n \text{ terms }$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow n = 10$

$\displaystyle \\$

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

Answer:

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

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

$\displaystyle \Rightarrow 3^n = 729$

$\displaystyle \Rightarrow 3^n = 3^6$

$\displaystyle \Rightarrow n = 6$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle n = 6$

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \Rightarrow n = 7$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Here we have } r = 3 a_n = 486 S_n = 728$

We k $\displaystyle \text{Now } a_n = ar^{n-1}$

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

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

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

Dividing ii) by i) we get

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

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

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

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

$\displaystyle \Rightarrow n = 6$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Let the first term be } a$ and the common ratio is $\displaystyle r$

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

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

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

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

Let $\displaystyle r^3 = x$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

Dividing ii) by i) we get

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

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

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

Substituting in i) we get $\displaystyle a = 1$

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Let the first term be } a \text{ and the common ratio be } r$

$\displaystyle \text{Given } a_2 = 24 \Rightarrow ar= 24$ … … … … … i)

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

Dividing ii) by i) we get

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Let the first term be } a \text{ and the common ratio be } r$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

For the first $\displaystyle n \text{ terms }$

$\displaystyle \text{Let the first term be } a$ and the common ratio is $\displaystyle r$ .

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

For next $\displaystyle n \text{ terms }$

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

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

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

Hence proved.

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } a , b$ are the roots of $\displaystyle x^2 - 3x + p = 0$

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

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

$\displaystyle \Rightarrow c+d = 12$ … … … … … $\displaystyle \text{iii) } \text{and } cd = q$ … … … … … iv)

$\displaystyle \text{Now } a, b, c, d$ are in G.P.

Let the common ratio be $\displaystyle r$ . Therefore

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

Substituting in i) & iii) we get

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

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

Dividing vi) by v) we get

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow n =10$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Here we have } a = 3 r = \frac{2^2}{2} = 2$

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{For } S_1, a = 1 \text{and } r = 1 \therefore S_1 = 1 + 1 + 1 + \ldots + 1 = n$

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

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

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

$\displaystyle \ldots$

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Let the first term be } a \text{ and the common ratio be } r$

Let there be $\displaystyle 2n \text{ terms }$ in the G.P.

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

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

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

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

$\displaystyle \Rightarrow 1 + r = 5$

$\displaystyle \Rightarrow r = 4$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Let the first term be } a \text{ and the common ratio be } r$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Let the given series be } a_1+a_2+a_3 +a_4 + \ldots + a_n$

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

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

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

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

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

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

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Class 11: Geometric Progression – Exercise 20.3

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