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?

Question

9366 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-26T16:52:30

Please refer to a Proof given in the 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.