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

Given that sum of the three terms$=\large\frac{39}{10}$

$\Rightarrow\:\large\frac{a}{r}$$+a+ar=\large\frac{39}{10}$.......(i)

Also given that their product $=1$

$\Rightarrow\:\large\frac{a}{r}$$\times a \times ar=1$

$\Rightarrow\:a^3=1$ or $a=1$

Substituting the value of $a$ in (i) we get

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

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

$\Rightarrow\:10r^2+10r+10=39r$

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

Solving this quadratic equation we get

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

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

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

$r=\large\frac{5}{2}$$\:\:or\:\:r=\large\frac{2}{5}$

Substituting the value of $a\:\:and\:\:r$ the three numbers are

$\large\frac{2}{5}$$,1,\large\frac{5}{2}$