Question

If \(a^2, b^2, c^2\) are in A.P., then \(b+c, c+a, a+b\) are in:

• A. AP
• B. GP
• C. HP
• D. None of these

Hint:

If squares are in A.P., then sums are in H.P. — very common exam result.

Answer 1

The Correct Option is C

Concept: If \(a^2, b^2, c^2\) are in A.P., then: \[ 2b^2 = a^2 + c^2 \] There is a standard result: \[ a^2, b^2, c^2 \text{ in A.P.} \;\Rightarrow\; \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in A.P.} \]

Step 1: Definition of H.P. A sequence is in H.P. if its reciprocals are in A.P.

Step 2: Apply result Since: \[ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in A.P.} \] \[ \Rightarrow b+c, \; c+a, \; a+b \text{ are in H.P.} \] Final: HP