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# How many terms of the series $6, 3, \large\frac{3}{2}$------ will make a sum of $\large\frac{765}{64}$.

$\begin{array}{1 1} 10 \\ 8 \\ 5\\ 12 \end{array}$

Explanation : In a GP
$S_{n}=\large\frac{a(r^n-1)}{r-1}\quad\;a=6,r=\large\frac{1}{2}$
$S_{n}=\large\frac{6(1-1/2^n)}{1-1/2}=\large\frac{765}{64}$
$1-(1/2)^n=\large\frac{765}{64*12}$
$(1/2)^n=1-\large\frac{765}{768}=\large\frac{3}{768}=\large\frac{1}{256}$
$(1/2)^n=\large\frac{1}{256}=(1/2)^8$
$n=8.$
edited Jan 24, 2014