Solution:

Let the three numbers be $a / r, a, ar$
According to the question
$\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=39 / 10 \ldots$ eq. (1)
$\mathrm{a} / \mathrm{r} \times \mathrm{a} \times \mathrm{ar}=1 \ldots$ eq. $(2)$
From eq. (2) we obtain,
$\begin{array}{l}
a^{3}=1 \\
a=1
\end{array}$
From eq. (1) we obtain,
$\left(a+a r+a r^{2}\right) / r=39 / 10$
$10 a+10 a r+10 a r^{2}=39 r \ldots$ eq. (3)
On substituting $a=1$ in 3 we obtain
$\begin{array}{l}
10(1)+10(1) r+10(1) r^{2}=39 r \\
10 r^{2}-29 r+10=0
\end{array}$
$10 r^{2}-25 r-4 r+10=0 \ldots$ eq. (4)
$\begin{array}{l}
5 r(2 r-5)-2(2 r-5)=0 \\
r=2 / 5 \text { or } 5 / 2
\end{array}$
The equation will now be,
$\begin{array}{l}
1 /(2 / 5), 1,1 \times 2 / 5 \text { or } 1 /(5 / 2), 1,1 \times 5 / 2 \\
5 / 2,1,2 / 5 \text { or } 2 / 5,1,5 / 2
\end{array}$
As a result, the three numbers are $2 / 5,1,5 / 2$