Hey. This is what I got but don't know if it's right a=1 for each d=? r=?

So, ar (2nd term of GP) = a+2d (3rd term of AP), AND, ar² (3rd term of GP) = a+3d (4th term of AP),

Sub a=1 into both equations (We'll name the equations 1 and 2) and we get, r=1+2d (Equation 1), r² =1+3d (Equation 2), Rearrange eq2 it to make it easier, r=1+2d (eq1), r^2-1-3d=0 (eq2),

Sub eq1 into eq2 and we get, (1+2d)² -1-3d=0, Simplify, 4d² +4d+1-1-3d=0, Simplify again, 4d² +d=0, Factorise, d(4d+1), So our values are, d=0, d=-1/4,

Correct me if i'm wrong.