Q.425. Arithmetic Progression

Question 425.
If 1/(a + b), 1/(b + c) and 1/(c + a) are in A.P., prove that
b^2, a^2 and c^2 are in A.P.

Answer 425.
If 1/(a+b), 1/(b+c) and 1/(c+a) are in A.P., then
1/(a+b) + 1/(c+a) = 2/(b+c)
=> (b+c)(c+a) + (a+b)(b+c) = 2(a+b)(c+a)
=> bc + c^2 + ab + ac + ab + b^2 + ac + bc
           = 2 (ac + bc + a^2 + ab)
=> b^2 + c^2 + 2(bc + ab + ac) = 2a^2 + 2(ab + bc + ca)
=> b^2 + c^2 = 2a^2
=> b^2, a^2 and c^2 are in A.P.

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