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}

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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] 

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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

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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]

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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

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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

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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

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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

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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

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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}

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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)

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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}}

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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}

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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.

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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.

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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}

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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

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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

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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

\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.

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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

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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}

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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)