Three consecutive A.P.: (a-d,\ a,\ a+d), sum (3a=6\implies a=2)

Increased by 1,4,15: [ 3-d,\ 6,\ 17+d ] in G.P.

[ 36=(3-d)(17+d)=51+3d-17d-d^2=51-14d-d^2 ] [ d^2+14d-15=0\implies(d+15)(d-1)=0 ] [ d=1\text{ or }-15 ]

For (d=1): numbers (1,2,3)

For (d=-15): (17,2,-13)

Answer: (1,2,3) (or (17,2,-13))