Question

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

Answer 1

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 ${(n+1)}^{th}$ to ${\left(2n\right)}^{th}$ term , sum of terms from ${(n+1)}^{th}$ to ${\left(2n\right)}^{th}$ term $=\frac{a_{n+1}{(1+r}^n)}{(1-r)}$
Thus required ratio $=\frac{a{(1+r}^n)}{(1-r)}\times \frac{(1-r)}{{ar}^n{(1-r}^n)}=\frac{1}{r^n}$
Thus the ratio of the sum of first $n$ terms of G.P. to the sum of terms from $(n+1{)}^{th}$ to ${\left(2n\right)}^{th}$ term is $\frac{1}{r^n}$