Question

Prove that a,b,c are in A.P. iff 1/bc,1/ca,1/ab are also in A.P.

Answer 1

  [tex] \frac{1}{ca} - \frac{1}{bc} = \frac{1}{ab} - \frac{1}{ca} as the common difference in an AP is same [/tex]

$\frac{bc-ca}{abc^2}= \frac{ca-ab}{a^2bc}=   \frac{c (b-a)}{abc^2} = \frac{a(c-b)}{a^2bc}$

No it remains b - a = c -b 
Hence this proves that a,b and c are also in AP.