The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152.

Let a be the first term and r be the common ratio of the G.P.

$\therefore S_{3}=a\left(\frac{r^{3}-1}{r-1}\right)$ and $S_{6}=a\left(\frac{r^{6}-1}{r-1}\right)$

Then, according to the question

$\frac{S_{3}}{S_{6}}=\frac{a\left(\frac{r^{3}-1}{r-1}\right)}{a\left(\frac{r^{6}-1}{r-1}\right)}$

$\Rightarrow \frac{125}{152}=\frac{r^{3}-1}{r^{6}-1}$$\Rightarrow \frac{125}{152}=\frac{r^{3}-1}{r^{6}-1}$

$\Rightarrow 125\left(r^{6}-1\right)=152\left(r^{3}-1\right)$

$\Rightarrow 125 r^{6}-125=152 r^{3}-152$

$\Rightarrow 125 r^{6}-152 r^{3}+27=0$

Now, let $r^{3}=y$

$\therefore 125 y^{2}-152 y+27=0$

Now, applying the quadatic formula

$y=\left\{\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right\}$

$\Rightarrow y=\left\{\frac{152 \pm \sqrt{9604}}{250}\right\}$

$\Rightarrow y=\left\{\frac{152+\sqrt{9604}}{250}\right\}$ or $\left\{\frac{152-\sqrt{9604}}{250}\right\}$

$\Rightarrow y=1$ or $\frac{27}{125}$

$\therefore r^{3}=1$ or $\quad r^{3}=\frac{27}{125}$

But, $r=1$ is not possible.

$\therefore r=\sqrt[3]{\frac{27}{125}}=\frac{3}{5}$