It is given that:

a^1/x = b^1/y = c^1/z

Let a^1/x = b^1/y = c^1/z = k

⇒ a^1/x = k

⇒ (a^1/x)^x = k^x…(Taking power of x on both sides.)

⇒ a^{1/x × x} = k^x

⇒ a = k^x

Similarly b = k^y

And c = k^z

It is given that a,b,c are in G.P.

⇒ b² = ac

Substituting values of a,b,c calculated above,we get:

⇒ (k^y)² = k^xk^z

⇒ k^2y = k^{x + z}

Comparing the powers we get,

2y = x + z

Which is the required condition for x,y,z to be in A.P.

Hence, proved that x,y,z, are in A.P.