Question

Three numbers a, b, c are in GP. and ax = by = c^z, then prove that x, z will be in H.P.

Answer 1

Let a^x = b^y = c^z = k

Then a = k^1/x, b = k^1/y, c = k^1/2 and a, b, c are in GP.

∴ b² = ac …(i)

Put the value of a, b, c in equation (i)

k^1/y = k^1/x, k^1/z or k^2/y = k^1/x+1/z

which is only possible, when

2/y = 1/x + 1/z

This, shows that 1/x, 1/y,1/z are in A.P. 

Then x, y, z will be in H.P.