It is given that:
a 1/x = b1/y = c1/z
Let a1/x = b1/y = c1/z = k
⇒ a 1/x = k
⇒ (a1/x) x = kx…(Taking power of x on both sides.)
⇒ a 1/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 = kxk z
⇒ 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.