Question

How many terms of the G.P., $3,\frac{3}{2},\frac{3}{4}...,$ are needed to give the sum $\frac{3069}{512}?$

Answer 1

Given: G.P. $3,\frac{3}{2},\frac{3}{4},...$ and $S_n=\frac{3069}{512}$

Here, first term $a=3$
common ratio $=r=\frac{\frac{3}{2}}{3}=\frac{1}{2}$
We know that, sum of $n$ terms of G.P. is given by
$S_n=\frac{a(1-r^n)}{1-r}$
$\frac{3069}{512}=\frac{3[1-{\left(\frac{1}{2}\right)}^n]}{1-\frac{1}{2}}$
$\Rightarrow \frac{3069}{512\times 6}=[1-{\left(\frac{1}{2}\right)}^n]$
$\Rightarrow {\left(\frac{1}{2}\right)}^n=(1-\frac{3069}{3072})$
$\Rightarrow {\left(\frac{1}{2}\right)}^n=\left(\frac{3072-3069}{3072}\right)$
$\Rightarrow {\left(\frac{1}{2}\right)}^n=\left(\frac{3}{3072}\right)$
$\Rightarrow {\left(\frac{1}{2}\right)}^n=\left(\frac{1}{1024}\right)$
$\Rightarrow {\left(\frac{1}{2}\right)}^n={\left(\frac{1}{2}\right)}^{10}$

On comparing of powers, we get
$n=10$
Hence, $10$ terms are needed to give the sum $\frac{3069}{512}$.