**Question:**

The first term of a GP is 27 , and its $8^{\text {th }}$ term is $\frac{1}{81}$. Find the sum of its first 10 terms.

**Solution:**

'Tn' represents the $n^{\text {th }}$ term of a G.P. series.

$\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$

$\Rightarrow \frac{1}{81}=27 \times r^{8-1}$

$\Rightarrow \frac{1}{81}=27 \times r^{7}$

$\Rightarrow \frac{1}{81} \div \frac{1}{27}=r^{7}$

$\Rightarrow \frac{1}{2187}=r^{7}$

$\Rightarrow\left(\frac{1}{3}\right)^{7}=r^{7}$

$\therefore \quad r=\frac{1}{3}$

Sum of a G.P. series is represented by the formula $\mathrm{Sn}=\mathrm{a} \frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}$ 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.

Here,

a = 27

$r=(\text { ratio between the } n \text { term and } n-1 \text { term })^{\frac{1}{3}}$

n = 10 terms

$\therefore \mathrm{S}_{\mathrm{n}}=27 \times \frac{1-\frac{1^{10}}{3}}{1-\frac{1}{3}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=27 \times \frac{1-\frac{1}{50049}}{\frac{2}{3}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=27 \times \frac{\frac{59048}{59049}}{\frac{2}{3}}$

$\Rightarrow S_{n}=27 \times \frac{39524}{19683}$

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{39524}{729}$