Solution:
Let the three numbers be $\mathrm{a} / \mathrm{r}, \mathrm{a}$, $ar$
According to the question,
$\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=38 \ldots$ eq.(1)
$\mathrm{a} / \mathrm{r} \times \mathrm{a} \times \mathrm{ar}=1728 \ldots$ eq.(2)
From eq.(2) we obtain,
$\begin{array}{l}
a^{3}=1728 \\
a=12
\end{array}$
From eq.(1) we obtain,
$\left(a+a r+a r^{2}\right) / r=38$
$\mathrm{a}+\mathrm{ar}+\mathrm{ar}^{2}=38 \mathrm{r} \ldots \text { eq. (3) }$
On substituting $a=12$ in eq.(3) we obtain
$12+12 r+12 r^{2}=38 r$
$12 r^{2}-26 r+12=0 \ldots \text { eq.(4) }$
By dividing eq.(4) by 2 we obtain
$\begin{array}{l}
6 r^{2}-13 r+6=0 \\
6 r^{2}-9 r-4 r+6=0 \\
3 r(3 r-3)-2(3 r-3)=0 \\
r=3 / 2 \text { or } r=2 / 3
\end{array}$
The eq. will now be
$\begin{array}{l}
12 /(3 / 2)=8 \text { or } \\
12 /(2 / 3)=18
\end{array}$
Therefore the terms are $8,12,18$
As a result, the three numbers are $8,12,18$