Three numbers form a GP. If the 3rd term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second term
of this AP is decreased by 8, a GP will be formed again, then the numbers will be

 let a, ar, ar²  are in GP  and a, ar, ar² - 64 are  in AP, we get

$\mathrm{a}\left({\mathrm{r}}^2-2\mathrm{r}+1\right)=64$      …….(i)

Again, a, ar -8, ar² - 64 are  in GP.

$\therefore (\mathrm{ar}-8{)}^2=\mathrm{a}\left({\mathrm{ar}}^2-64\right)$

$\Rightarrow \mathrm{a}(16\mathrm{r}-64)=64$     …..(ii)

On solving  Eqs.(i) and (ii), we get r =5,a =4

Thus,  required numbers  are 4, 20, 100.