Given that a^x = b^y = c^z = d^u and a,b,c,d are in G.P, show that x,y,z,u are in H.P?

1 Answer
Aug 26, 2017

Please refer to a Proof given in the Explanation.

Explanation:

Given that, a^x=b^y=c^z=d^u.

Suppose that, a^x=b^y=c^z=d^u=k.

:. a=k^(1/x), b=k^(1/y), c=k^(1/z), and, d=k^(1/u).........(1).

Now, a, b, c, and, d are in G.P.

:. b/a=c/b=d/c.

By (1)," then, "k^(1/y)/k^(1/x)=k^(1/z)/k^(1/y)=k^(1/u)/k^(1/z), or,

k^(1/y-1/x)=k^(1/z-1/y)=k^(1/u-1/z),

rArr (1/y-1/x)=(1/z-1/y)=(1/u-1/z),

rArr 1/x, 1/y, 1/z, 1/u, are in A.P.

rArr x,y,z,u are in H.P.

Hence, the Proof.