Find the sum of the following series to infinity:

(i) In the given G.P., first term, $a=1$

and common ratio, $r=-\frac{1}{3}$

Hence, the sum $S$ to infinity is given by $S=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-\left(-\frac{1}{3}\right)}=\frac{3}{4}$.

(ii) In the given G.P., first term, $a=8$

and common ratio, $r=\frac{1}{\sqrt{2}}$

Hence, the sum $S$ to infinity is given by $S=\frac{a}{1-r}=\frac{8}{1-\frac{1}{\sqrt{2}}}=(2+\sqrt{2})$.

(iii) We have:

$\frac{2}{5}+\frac{3}{5^{2}}+\frac{2}{5^{3}}+\frac{3}{5^{4}}$

$=\left(\frac{2}{5}+\frac{2}{5^{3}}+\ldots \infty\right)+\left(\frac{3}{5^{2}}+\frac{3}{5^{4}}+\ldots\right)$

$=\left(\right.$ An infinite G.P. with $a=\frac{2}{5}$ and $\left.r=\frac{1}{25}\right)+\left(\right.$ An infinite G.P. with $a=\frac{3}{25}$ and $\left.r=\frac{1}{25}\right)$

$=\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right)+\left(\frac{\frac{3}{5}}{1-\frac{1}{25}}\right)$

$=\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\left(\frac{\frac{3}{5}}{\frac{24}{25}}\right)$

$=\left(\frac{10}{24}+\frac{3}{24}\right)$

$=\frac{13}{24}$