Show that the ratio of the sum of the first $n$ terms of a $G.P.$ to the sum of the terms from the $(n+1)^{th}$ to the $(2n)^{th}$ term is $\frac{1}{r^{n}}$.

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The sum of the first $n$ terms is given by $S_n = \frac{a(1-r^n)}{1-r}$.
The terms from the $(n+1)^{th}$ to the $(2n)^{th}$ term form a $G.P.$ with the first term $a_{n+1} = ar^n$ and the number of terms equal to $n$.
The sum of these terms is $S' = \frac{a_{n+1}(1-r^n)}{1-r} = \frac{ar^n(1-r^n)}{1-r}$.
The ratio of the sum of the first $n$ terms to the sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ is:
$\text{Ratio} = \frac{\frac{a(1-r^n)}{1-r}}{\frac{ar^n(1-r^n)}{1-r}} = \frac{a(1-r^n)}{1-r} \times \frac{1-r}{ar^n(1-r^n)} = \frac{1}{r^n}$.
Thus,the ratio is $\frac{1}{r^n}$.