Step 1

Given that the G.P. has $2n$ terms.

Let the first $2n$ terms of the G.P be

$(a+ar+ar^2+......ar^{n-1})+(ar^n+ar^{n+1}+....ar^{2n-1})$

Let the sum of first $n$ terms be $S_n$,

Sum of first $2n$ terms be $S_{2n}$ and

Sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term be $S$

We know that the sum of first $n$ terms of a G.P$=S_n=a.\large\frac{1-r^n}{1-r}$

$\Rightarrow\:$The sum of first $2n$ terms of the G.P$=S_{2n}=a.\large\frac{1-r^{2n}}{1-r}$

$\therefore\:S_{2n}=(a+ar+ar^2+......ar^{n-1})+(ar^n+ar^{n+1}+....ar^{2n-1})$

$\Rightarrow\:S_{2n}=S_n+S$

$\Rightarrow\:a.\large\frac{1-a^{2n}}{1-r}$$=a.\large\frac{1-r^n}{1-r}$$+S$

$\Rightarrow\:$Sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term $=S=$

$(a.r^{n}+a.r^{n+1}+........a.r^{2n-1})=S=a.\large\frac{1-a^{2n}}{1-r}$$-a.\large\frac{1-r^n}{1-r}$

Taking $a\:\;and\:\:(1-r)$ common we get

$\Rightarrow\:S=\large\frac{a}{1-r}$$.(1-r^{2n}-1+r^n)$

$\Rightarrow\:S=\large\frac{a}{1-r}$$.(r^n-r^{2n})$

Taking $r^n$ common we get

$\Rightarrow\:S=\large\frac{ar^n}{1-r}$$.(1-r^{n})$

Step 2

The ratio of the first $n$ terms of a G.P. to the sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term =

$S_n\: : \:S=a.\large\frac{1-r^n}{1-r}\: :\: \frac{a.r^n}{1-r}.$$(1-r^n)$

Cancelling the terms $a,\:\:(1-r^n)\:\:and\:\:(1-r)$ on both the sides we get

$\Rightarrow\:S_n\: :\:S=1\: :\:r^n$

Hence Proved.