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 \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)