B=A+d=Ar; C=A+2d=Ar^2, where d is the common difference and r is the common ratio.
Therefore, Ar-A=d, and A=d/(r-1) and Ar^2-A=2d, A=2d/(r^2-1).
And d/(r-1)=2d/(r^2-1), from which 1/(r-1)=r^2-1=2(r-1); r^2-1-2r+2=0 or (r-1)^2=0 and r=1 and d=0.
So A=B=C. But this is the trivial solution.
If A, B and C are the first three terms but not necessarily in that order, we have 2d/(r-1)=d/(r^2-1) where the order is A, C, B. This becomes 2/(r-1)=1/(r^2-1).
Assume r≠1, 2=1/(r+1) because r^2-1=(r-1)(r+1) and r-1 is a common factor. So r+1=1/2 and r=-1/2. From this value of r we have A+2d=rA=-A/2 and d=-3A/4 which also satisfies A+d=A/4. So the three terms are A, A-3A/4, A-3A/2 which is A, A/4, -A/2 for the AP and A, -A/2, A/4 for the GP. A can be any number. Put A=4 for example: AP is 4, 1, -2 and GP is 4, -2, 1. So r=-1/2 and d=-3A/4.