Question

If \((b+c), (c+a), (a+b)\) are in harmonic progression, then \(a^2, b^2, c^2\) are in

• A. AP
• B. GP
• C. HP
• D. Both AP and HP

Hint:

Substitute S = a+b+c so each bracket becomes S minus a variable, then apply the AP condition to the reciprocals.

Answer 1

The Correct Option is A

Step 1: Rewrite the terms using S = a + b + c.
b+c = S-a, c+a = S-b, a+b = S-c, so \(\frac{1}{S-a}, \frac{1}{S-b}, \frac{1}{S-c}\) are in AP.
Step 2: Apply the AP condition.
\(\frac{2}{S-b} = \frac{1}{S-a} + \frac{1}{S-c} = \frac{(S-c)+(S-a)}{(S-a)(S-c)} = \frac{2S-a-c}{(S-a)(S-c)}\).
Since \(2S-a-c = S+b\) (because \(S-a-c=b\)), this gives \(2(S-a)(S-c) = (S-b)(S+b) = S^2-b^2\).
Step 3: Expand and simplify.
\(2(S-a)(S-c) = 2S^2 - 2S(a+c) + 2ac\). Using \(a+c = S-b\): this becomes \(2Sb + 2ac\).
So \(2Sb+2ac = S^2-b^2\), i.e. \(S^2 = 2Sb+2ac+b^2\).
Step 4: Substitute \(S=a+b+c\) and expand fully.
\((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\) and \(2Sb = 2ab+2b^2+2bc\).
Equating: \(a^2+b^2+c^2+2ab+2bc+2ca = 2ab+2bc+2ac+3b^2\), which reduces to \(a^2+c^2 = 2b^2\).
This is exactly the condition for \(a^2, b^2, c^2\) to be in AP.\[\boxed{a^2,\ b^2,\ c^2\ are\ in\ AP}\]