The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92.

(a) 1/2

Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots, \infty$.

$S_{\infty}=4$

$\Rightarrow \frac{a}{1-r}=4$   ...(i)

Also, sum of the cubes, $S_{1}=92$

$\Rightarrow \frac{a^{3}}{\left(1-r^{3}\right)}=92$    ...(ii)

Putting the value of $a$ from $(\mathrm{i})$ to $(\mathrm{ii})$ :

$\Rightarrow \frac{(4(1-r))^{3}}{\left(1-r^{3}\right)}=92$

$\Rightarrow \frac{64(1-r)^{3}}{\left(1-r^{3}\right)}=92$

$\Rightarrow \frac{(1-r)^{3}}{(1-r)\left(1+r+r^{2}\right)}=\frac{92}{64}$

$\Rightarrow \frac{(1-r)^{2}}{\left(1+r+r^{2}\right)}=\frac{23}{16}$

$\Rightarrow 16\left(1-2 r+r^{2}\right)=23\left(1+r+r^{2}\right)$

$\Rightarrow 7 r^{2}+55 r+7=0$

Using the quadratic formula:

$\Rightarrow r=\frac{-55+\sqrt{55^{2}-4 \times 7 \times 7}}{2 \times 7}$

$\Rightarrow r=\frac{-55+\sqrt{55^{2}-14^{2}}}{14}$

$\Rightarrow r=\frac{-55+\sqrt{2829}}{14}$

Disclaimer: None of the given options are correct. This solution has been created according to the question given in the book.