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 termof 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 getar2−2r+1=64      …….(i)Again, a, ar -8, ar2 - 64 are  in GP.∴ (ar−8)2=aar2−64⇒ a(16r−64)=64     …..(ii)On solving  Eqs.(i) and (ii), we get r =5,a =4Thus,  required numbers  are 4, 20, 100.