Mathematics - Sequence & Series Question with Solution | TestHub

The series is a G.P. with common ratio $=\left(\frac{1-3\mathrm{x}}{1+3\mathrm{x}}\right)$ and $|\mathrm{r}|=\left|\frac{1-3\mathrm{x}}{1+3\mathrm{x}}\right|$ is less than 1 since $\mathrm{x}$ is positive ${\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{1+3\mathrm{x}}}{1-\left\{-\left(\frac{1-3\mathrm{x}}{1+3\mathrm{x}}\right)\right\}}=\frac{1}{2}$