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

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

Toolbox:
• 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

edited Nov 19, 2013