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

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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

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