Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive,

Let $r$ be the common ratio of the given G.P.

$\therefore a=4$

Sum of the geometric ifinite series:

$S_{\infty}=4+4 r+4 r^{2}+\ldots \infty$

Now, $S_{\infty}=\frac{4}{1-r} \quad \ldots \ldots$ (i)

The difference between the third and fifth term is $\frac{32}{81}$.

$a_{3}-a_{5}=\frac{32}{81}$

$\Rightarrow 4 r^{2}-4 r^{4}=\frac{32}{81}$

$\Rightarrow 4\left(r^{2}-r^{4}\right)=\frac{32}{81}$

$\Rightarrow 81 r^{4}-81 r^{2}+8=0$    ....(ii)

Now, let $r^{2}=y$

Let us put this in (ii).

$\therefore 81 r^{4}-81 r^{2}+8=0$

$\Rightarrow 81 y^{2}-81 y+8=0$

$\Rightarrow 81 y^{2}-72 y-9 y+8=0$

$\Rightarrow 9 y(9 y-1)-8(9 y-1)=0$

$\Rightarrow(9 y-8)(9 y-1)$

$\Rightarrow y=\frac{1}{9}, \frac{8}{9}$

Putting $y=r^{2}$, we get $\mathrm{r}=\frac{1}{3}$ and $\frac{2 \sqrt{2}}{3}$

Substituting $\mathrm{r}=\frac{1}{3}$ and $\frac{2 \sqrt{2}}{3}$ in (i):

$\mathrm{S}_{\infty}=\frac{4}{1-\frac{1}{3}}=\frac{12}{2}=6$

Similarly, $\mathrm{S}_{\infty}=\frac{4}{1-\frac{2 \sqrt{2}}{3}}=\frac{12}{3-2 \sqrt{2}}$

$\therefore S_{\infty}=6, \frac{12}{3-2 \sqrt{2}}$