SERIES

edited

HOW DID U GET THAT>

YES

This is simple....................

X^A = Z^C
X = Z^(C/A)
Z^C = (X.Z)^B/2 = [ Z^(C/A + 1) ]^B/2 = Z^{(A+C)B/2A}

C = (A+C)B/2A
AC/(A+C) = B/2
(A+C)/AC = 2/B

(1/A + 1/C)/2 = 1/B

Hence 1/A,1/B & 1/C are in A.P.

NICE.................ACTUALLY THIS ONE WAS NOT MY DOUBT :) GOOD JOB

THE KEY IDEA WAS THAT............ONCE U REALIZE THAT U CAN EXPLICITY REDUCE GIVEN INFO TO ONLY ONE VARIABLE Z//////////IT WUD BE EASY OFF............. :)