Solution:

Given that,
The sum of G.P of 3 terms is 125
Using the formula,
The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
$125=\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right) /(\mathrm{r}-1)$
$125=\mathrm{a}\left(\mathrm{r}^{3}-1\right) /(\mathrm{r}-1) \ldots$ eq. (1)
So now,
The sum of G.P of $6$ terms is $152$
Using the formula,
The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
$152=\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right) /(\mathrm{r}-1)$
$152=a\left(r^{6}-1\right) /(r-1) \ldots$ eq. (2)
On dividing the eq. (i) by eq. (ii) we obtain,
$\begin{array}{l}
125 / 152=\left[a\left(r^{3}-1\right) /(r-1)\right] /\left[a\left(r^{6}-1\right) /(r-1)\right] \\
125 / 152=\left(r^{3}-1\right) /\left(r^{6}-1\right) \\
125 / 152=\left(r^{3}-1\right) /\left[\left(r^{3}-1\right)\left(r^{3}+1\right)\right] \\
125 / 152=1 /\left(r^{3}+1\right) \\
125\left(r^{3}+1\right)=152 \\
125 r^{3}+125=152 \\
125 r^{3}=152-125 \\
125 r^{3}=27 \\
r^{3}=27 / 125 \\
r^{3}=3^{3} / 5^{3} \\
r=3 / 5
\end{array}$
Therefore, the common ratio is $3 / 5$