Question:
Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^{n}}$.
Solution:
Let a be the first term and r be the common ratio of the G.P.
Sum of the first $n$ terms of the series $=a_{1}+a_{2}+a_{3}+\ldots+a_{n}$
Similarly, sum of the terms from $(n+1)^{\text {th }}$ to $2 n^{\text {th }}$ term $=a_{n+1}+a_{n+2}+\ldots+a_{2 n}$
$\therefore$ Required ratio $=\frac{a_{1}+a_{2}+a_{3}+\ldots+a_{n}}{a_{n+1}+a_{n+2}+\ldots+a_{2 n}}$
$=\frac{a+a r+\ldots+a r^{n-1}}{a r^{n}+a r^{n+1}+\ldots+a r^{2 n-1}}$
$=\frac{a\left(\frac{1-r^{n}}{1-r}\right)}{a r^{n}\left(\frac{1-r^{n}}{1-r}\right)}$
$=\frac{1}{r^{n}}$