The 4 th and 7 th terms of a G.P. are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of $n$ terms of the G.P.
Let a be the first term and r be the common ratio of the G.P.
$\therefore a_{4}=\frac{1}{27}$
$\Rightarrow a r^{4-1}=\frac{1}{27}$
$\Rightarrow a r^{3}=\frac{1}{27}$
$\Rightarrow\left(a r^{3}\right)^{2}=\frac{1}{27^{2}}$
$\Rightarrow a^{2} r^{6}=\frac{1}{729}$
$\Rightarrow a r^{6}=\frac{1}{729 a}$ ...(i)
Similarly, $\mathrm{a}_{7}=\frac{1}{729}$
$\Rightarrow a r^{7-1}=\frac{1}{729}$
$\Rightarrow a r^{6}=\frac{1}{729}$
$\Rightarrow a r^{6}=\frac{1}{729 \mathrm{a}}$ [From (i)]
Putting this in $a_{4}=\frac{1}{27}$
$\Rightarrow a r^{3}=\frac{1}{3^{3}}$
$\Rightarrow r^{3}=\frac{1}{3^{3}}$
$\therefore r=\frac{1}{3}$
Now, sum of $n$ terms of the G.P., $S_{n}=a\left(\frac{r^{n}-1}{r-1}\right)$
$\Rightarrow S_{n}=1\left(\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}\right)$
$\Rightarrow S_{n}=\frac{3}{2}\left(1-\frac{1}{3^{n}}\right)$