Question:hard

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

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Substitute S = a+b+c so each bracket becomes S minus a variable, then apply the AP condition to the reciprocals.
Updated On: Jul 21, 2026
  • AP
  • GP
  • HP
  • Both AP and HP
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The Correct Option is A

Solution and Explanation

Step 1: Pick convenient numbers satisfying the given HP condition.
Let b+c = 2, a+b = 3. For an HP, \(\frac{2}{c+a} = \frac{1}{b+c}+\frac{1}{a+b} = \frac{1}{2}+\frac{1}{3} = \frac{5}{6}\), so \(c+a = \frac{12}{5} = 2.4\).
Step 2: Solve for a, b, c.
Adding all three: \(2(a+b+c) = 2+2.4+3 = 7.4\), so \(a+b+c = 3.7\).
Then \(a = 3.7-2 = 1.7\), \(b = 3.7-2.4 = 1.3\), \(c = 3.7-3 = 0.7\).
Step 3: Check which progression \(a^2,b^2,c^2\) follow.
\(a^2=2.89,\ b^2=1.69,\ c^2=0.49\).
Differences: \(b^2-a^2 = 1.69-2.89 = -1.20\) and \(c^2-b^2=0.49-1.69=-1.20\). The two differences are equal.
Step 4: Conclude.
Equal consecutive differences mean \(a^2,b^2,c^2\) are in Arithmetic Progression (this matches the general proof, confirming the result for any valid a, b, c).\[\boxed{a^2,\ b^2,\ c^2\ are\ in\ AP}\]
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