Question

The number of terms of G.P. $3,\frac{3}{2},\frac{3}{4},\cdots$required to give the sum $\frac{3069}{512}$ is

Answer 1

$a=3,r=\frac{1}{2}$
So, $S_n=\frac{3[1-(1/2{)}^n]}{1-\frac{1}{2}}=6[1-\frac{1}{2^n}]$
Now $6[1-\frac{1}{2^n}]=\frac{3069}{512}$
$\Rightarrow 1-\frac{1}{2^n}=\frac{1023}{1024}\Rightarrow \frac{1}{2^n}=\frac{1}{1024}=\frac{1}{2^{10}}\therefore n=10$