11This 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.
7 Answers
11@mrunal.......
how do u get A=B/2=C
i think the ques. is X^A = (X.Z)^B/2 = Z^C
11
1NICE.................ACTUALLY THIS ONE WAS NOT MY DOUBT :) GOOD JOB
1THE KEY IDEA WAS THAT............ONCE U REALIZE THAT U CAN EXPLICITY REDUCE GIVEN INFO TO ONLY ONE VARIABLE Z//////////IT WUD BE EASY OFF............. :)