**Question:**

Find the sum of the GP :

$1-\frac{1}{2}+\frac{1}{4}=\frac{1}{8}+\ldots$ to 9 terms

**Solution:**

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

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 = 1

$r=($ ratio between the $n$ term and $n-1$ term $)-\frac{1}{2} \div 1=-\frac{1}{2}$

n = 9 terms

$\therefore \mathrm{S}_{\mathrm{n}}=1 \times \frac{1-\frac{-1^{9}}{2}}{1-\left(\frac{-1}{2}\right)}$

$\Rightarrow S_{\mathrm{n}}=\frac{1+\frac{1}{512}}{1+\frac{1}{2}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\frac{513}{2}}{\frac{3}{2}}$

$\therefore \mathrm{S}_{\mathrm{n}}=171$