Question

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

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

Hint:

Convert the HP condition on \((b+c),(c+a),(a+b)\) into an AP condition on their reciprocals, substitute \(s=a+b+c\), and simplify.

Answer 1

The Correct Option is A

Let \(s = a+b+c\), so that \(b+c = s-a\), \(c+a = s-b\), and \(a+b = s-c\).

Step 1: Write the HP condition.
Since \((s-a), (s-b), (s-c)\) are in HP, their reciprocals are in AP, which means:
\[\frac{2}{s-b} = \frac{1}{s-a} + \frac{1}{s-c}\]

Step 2: Cross-multiply.
\[2(s-a)(s-c) = (s-b)\big[(s-a)+(s-c)\big]\]
Now, \((s-a)+(s-c) = 2s - a - c\). Since \(a+c = s-b\), this becomes \(2s-(s-b) = s+b\). So:
\[2(s-a)(s-c) = (s-b)(s+b) = s^2 - b^2\]

Step 3: Expand the left side.
\[2(s-a)(s-c) = 2\big[s^2 - s(a+c) + ac\big] = 2\big[s^2 - s(s-b) + ac\big] = 2\big[sb + ac\big] = 2sb + 2ac\]
So the equation becomes:
\[2sb + 2ac = s^2 - b^2 \quad \Rightarrow \quad s^2 - 2sb - b^2 = 2ac\]

Step 4: Substitute \(s = a+b+c\).
\[s^2 = a^2+b^2+c^2+2ab+2bc+2ca\]\[s^2 - 2sb = a^2+b^2+c^2+2ab+2bc+2ca - 2b(a+b+c) = a^2 - b^2 + c^2 + 2ca\]
So:
\[a^2 - b^2 + c^2 + 2ca - b^2 = 2ac \quad \Rightarrow \quad a^2 - 2b^2 + c^2 + 2ca - 2ac = 0\]
Since \(2ca - 2ac = 0\), this simplifies beautifully to:
\[a^2 - 2b^2 + c^2 = 0 \quad \Rightarrow \quad a^2 + c^2 = 2b^2\]

Step 5: Interpret the result.
The condition \(a^2+c^2 = 2b^2\) is exactly the defining condition for \(a^2, b^2, c^2\) to be in Arithmetic Progression (the middle term is the average of the other two: \(b^2 = \frac{a^2+c^2}{2}\)).

So \(a^2, b^2, c^2\) are in AP, which is option (a).