Answer : It is given that:

a1/x = b1/y = c1/z

Let a1/x = b1/y = c1/z = k

⇒ a1/x = k

⇒ (a1/x)x = kx…(Taking power of x on both sides.)

⇒ a1/x × x = kx

⇒ a = kx Similarly b = ky And c = kz

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

⇒ b2 = ac

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

⇒ (ky)2 = kxkz

⇒ k2y = kx + 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.