The sum of three numbers in a G.P is $\frac{13}{12}$  and their product is $-1$ , then the numbers are

Let the numbers $\frac{a}{r},a,ar$

$\frac{a}{r}+a+ar=\frac{13}{12}\to \left(1\right)$

$\frac{a}{r}.a.ar=-1$

$a^3=-1$

$a=-1$

$\frac{-1}{r}-1-r=\frac{13}{12}$

$\frac{-1-r-r^2}{r}=\frac{13}{12}$

$13r=-12-12r-12r^2$

$12r^2+25r+12=0$

$12r^2+16r+9r+12=0$

$4r\left(3r+4\right)+3\left(3r+4\right)=0$

$\left(3r+4\right)\left(4r+3\right)=0$

$r=\frac{-4}{3}$  (or) $r=\frac{-3}{4}$

Numbers are $\frac{-1}{\left(\frac{-4}{3}\right)},-1,\left(-1\right)\left(\frac{-4}{3}\right)$

$\frac{3}{4},-1,\frac{4}{3}$   (or)  $\frac{4}{3},-1,\frac{3}{4}$