# Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

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

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

Sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term $=\frac{\mathrm{a}_{\mathrm{n}+1}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{(1-\mathrm{r})}$

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

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 $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is ${ }^{\frac{1}{n^{n}}}$.