Given: If a, b, c are in G.P. and a^1/x = b^1/y = c^1/z, prove that x, y, z 

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

Then, a = k^x, b = k^y, and c = k^z.     ...(1)

Since a, b, c are in G.P., therefore b² = a·c.     ...(2)

Using (1) in (2), we get:

k^2y = k^x+z

⇒ 2y = x + z

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