**Question:**

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

**Solution:**

Sum of a G.P. series is represented by the formula $S_{n}=a \frac{r^{n}-1}{r-1}$ when r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Thus, the sum of the first n terms of the G.P. series is, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

Sum of $(n+1)^{\text {th }}$ term to $2 n^{\text {th }}$ term

$=$ Sum of the first $2 \mathrm{n}^{\text {th }}$ term - the sum of $1^{\text {st }}$ term to $\mathrm{n}^{\text {th }}$ term

$=a \frac{r^{2 n}-1}{r-1}-a \frac{r^{n}-1}{r-1}$

$=\frac{\left(a r^{2 n}-a\right)-\left(a r^{n}-a\right)}{r-1}$

$=\frac{a r^{2 n}-a-a r^{n}+a}{r-1}$

$=\frac{\operatorname{ar}^{\mathrm{n}}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

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

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

[Cancelling out the common factors from the numerator and denominator $\Rightarrow a,(r-1)$, $\left.\left(r^{n}-1\right)\right]$

$=\frac{1}{r^{n}}$

Hence Proved.

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