Solution:

According to the question
$\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=65 \ldots$ equation (1)
$a / r \times a \times a r=3375 \ldots$ equation (2)
From eq.(2) we obtain,
$\begin{array}{l}
a^{3}=3375 \\
a=15
\end{array}$
From eq.(1) we obtain,
$\left(a+a r+a r^{2}\right) / r=65$
$a+a r+a r^{2}=65 r \ldots$ eq.(3)
By substituting $a=15$ in eq.(3) we obtain
$15+15 r+15 r^{2}=65 r$
$15 r^{2}-50 r+15=0 \ldots \text { eq. (4) }$
On dividing eq.(4) by 5 we obtain
$\begin{array}{l}
3 r^{2}-10 r+3=0 \\
3 r^{2}-9 r-r+3=0 \\
3 r(r-3)-1(r-3)=0 \\
r=3 \text { or } r=1 / 3
\end{array}$
The eq. will now be
$15 / 3,15,15 \times 3$ or
$15 /(1 / 3), 15,15 \times 1 / 3$
So the terms are $5,15,45$ or $45,15,5$
As a result, the three numbers are $5,15,45$