In a G.P., the product of first four terms is 4 and the second term is the reciprocal of the fourth term. The sum of infinite terms of the G.P can be

$a,ar,a{r}^2,a{r}^3$  are four terms in G.P

$\left(a\right).\left(ar\right).\left(a{r}^2\right).\left(a{r}^3\right)=4$

$a^4{r}^6=4\to \left(1\right)$

and also $ar=\frac{1}{a{r}^3}$

$a^2{r}^4=1$

$a^4{r}^8=1\to \left(2\right)$

$\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{a^4{r}^8}{a^4{r}^6}=\frac{1}{4}$

$\Rightarrow r^2=\frac{1}{4}$

$\Rightarrow r=\pm \frac{1}{2}$

If $r=\frac{1}{2}$

From (2),$a^4{\left(\frac{1}{2}\right)}^8=1$

$a^4=2^8$

$a=4$ , $-4$  also satisfy the equation $a^4=2^8$

If $a=4$                     $a=-4$

$S_{\infty }=\frac{a}{1-r}$                 $S_{\infty }=\frac{a}{1-r}$

$=\frac{4}{1-\frac{1}{2}}$                      $=\frac{-4}{1-\frac{1}{2}}$

$=8$                            $=-8$

If $r=-\frac{1}{2}$  

If $a=4$                    $a=-4$

$S_{\infty }=\frac{a}{1-r}$                  $S_{\infty }=\frac{a}{1-r}$

$=\frac{4}{1+\frac{1}{2}}$                      $=\frac{-4}{1+\frac{1}{2}}$

$=\frac{8}{3}$                            $=-\frac{8}{3}$