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Find the least number of terms for the series 16, 24, 36,.... such that the sum of all terms of the series exceeds 2000

$(a)\;11\qquad(b)\;3\qquad(c)\;13\qquad(d)\;15$

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  • Sum of n terms of a GP is $S_{n} = \large \frac{a ( r^{n} - 1)} {r - 1} $
Looking at the series, we find that it is a GP with $ a = 16 $ and $ r = \Large\frac{3}{2} $
Given that $ S_{n} > 2000 $
 
Sum of n terms of a GP is $S_{n} = \large \frac{a ( r^{n} - 1)} {r - 1} $
 
$16 \Large\frac {((\Large\frac {3}{2})^n - 1)} {\Large\frac {3}{2} -1} > 2000$
 
$ (1.5)^{n} > 63.5 $
 
$ n > \frac {\log_{10}63.5} {\log_{10} 1.5}$
 
$ \implies n > 10.3 $
n = 11 as the least natural number is required to be found

Therefore, the sum of 11 terms of the series 16, 24, 36,  ... will exceed 2000

answered Nov 19, 2013 by harini.tutor
edited Nov 19, 2013 by harini.tutor
 

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