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

Step 1: Understanding the Concept:
Terms are in Arithmetic Progression (AP) if their common differences are equal. Terms are in Harmonic Progression (HP) if their reciprocals are in AP.
Step 3: Detailed Explanation:
1. Given \(a^{2}, b^{2}, c^{2}\) are in AP.
By property, \(b^{2} - a^{2} = c^{2} - b^{2}\).
2. Add \((ab + bc + ca)\) to each term:
\[ a^{2} + ab + bc + ca, \quad b^{2} + ab + bc + ca, \quad c^{2} + ab + bc + ca \text{ are also in AP} \]
3. Factorize the expressions:
\[ (a + b)(a + c), \quad (b + c)(b + a), \quad (c + a)(c + b) \text{ are in AP} \]
4. Divide all terms by \((a + b)(b + c)(c + a)\):
\[ \frac{(a+b)(a+c)}{(a+b)(b+c)(c+a)}, \quad \frac{(b+c)(b+a)}{(a+b)(b+c)(c+a)}, \quad \frac{(c+a)(c+b)}{(a+b)(b+c)(c+a)} \text{ are in AP} \]
\[ \frac{1}{b+c}, \quad \frac{1}{c+a}, \quad \frac{1}{a+b} \text{ are in AP} \]
5. Since the reciprocals are in AP, the terms \(b + c, c + a, a + b\) must be in Harmonic Progression (HP).
Step 4: Final Answer:
The terms are in HP.