SOLUTION: If b+c-a/a, c+a-b/b, a+b-c/c are in A.P. Prove that 1/a,1/b,1/c are also in A.P

Algebra ->  Sequences-and-series -> SOLUTION: If b+c-a/a, c+a-b/b, a+b-c/c are in A.P. Prove that 1/a,1/b,1/c are also in A.P      Log On


   

Question 914044: If b+c-a/a, c+a-b/b, a+b-c/c are in A.P.
Prove that 1/a,1/b,1/c are also in A.P
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
b+c-a/a, c+a-b/b, a+b-c/c in AP

+%28c%2Ba-b%29%2Fb+-+%28b%2Bc-a%29%2Fa+=+%28a%2Bb-c%29%2Fc+-+%28c%2Ba-b%29%2Fb

+%28a%28c%2Ba-b%29+-+b%28b%2Bc-a%29%29%2Fab+=+%28b%28a%2Bb-c%29-c%28c%2Ba-b%29%29%2Fbc

%28ac%2Ba%5E2-ab-b%5E2-bc%2Bab%29%2Fab+=+%28ab%2Bb%5E2-bc-c%5E2-ac%2Bbc%29%2Fbc


%28a%5E2-b%5E2%2Bac-bc%29%2Fab+=+%28+b%5E2-c%5E2%2Bab-ac%29%2Fbc

%28%28a%2Bb%29%28a-b%29%2Bc%28a-b%29%29%2Fab=%28%28b%2Bc%29%28b-c%29%2Ba%28b-c%29%29%2Fbc


%28%28a-b%29%28a%2Bb%2Bc%29%29%2Fab=+%28%28b-c%29%28a%2Bb%2Bc%29%29%2Fbc

%28a-b%29%2Fa=+%28b-c%29%2Fc

ac-bc = ab-ac
/abc
1/b -1/a = 1/c-1/b
Hence they are in AP