Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

Sum of first $n$ terms $=\frac{a\left(1-r^{n}\right)}{(1-r)}$

Since there are $n$ terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term,

Sum of terms from $(n+1)^{t h}$ to $(2 n)^{th}$ term

$S_{n}=\frac{a_{n+1}\left(1-r^{n}\right)}{1-r}$

$a^{n+1}=a r^{n+1-1}=a r^{n}$

Thus, required ratio $=\frac{a\left(1-r^{n}\right)}{(1-r)} \times \frac{(1-r)}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}}$

Thus, the ratio of the sum of first $n$ terms of a $G.P.$ to the sum of terms from term is $\frac{1}{r^{n}}$