Let $\frac{a}{r}, a,$ ar be the first three terms of the $G.P.$

$\frac{a}{r}+a+a r=\frac{39}{10}$        ……….$(1)$

$\left(\frac{a}{r}\right)(a)(a r)=1$         ………$(2)$

From $(2),$ we Obtain $a^{3}=1$

$\Rightarrow a=1$ (Considering real roots only)

Substituting $a=1$ in equation $(1),$ we obtain

$\frac{1}{r}+1+r=\frac{39}{10}$

$\Rightarrow 1+r+r^{2}=\frac{39}{10} r$

$\Rightarrow 10+10 r+10 r^{2}-39 r=0$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$

$\Rightarrow(5 r-2)(2 r-5)=0$

$\Rightarrow r=\frac{2}{5}$ or $\frac{5}{2}$

Thus, the three terms of $G.P.$ are $\frac{5}{2}, 1$ and $\frac{2}{5}$