Question:hard

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

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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.
Updated On: Jul 20, 2026
  • AP
  • GP
  • HP
  • Both AP and HP
  • None of these
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The Correct Option is A

Solution and Explanation

Instead of grinding through the general algebra, this type of progression question can be checked quickly by testing it on a set of numbers that actually satisfies the given condition, and then confirming the general proof matches.

Take $a=1, b=2, c=3$ as a trial (chosen freely, not because they satisfy HP - this is just to understand structure). Sum $s=6$, so $b+c=5, c+a=4, a+b=3$. For HP we'd need $1/5, 1/4, 1/3$ to be in AP, and $1/4-1/5 = 0.05$ while $1/3-1/4 = 0.0833$, so these particular numbers don't satisfy HP - meaning $a,b,c$ can't be picked arbitrarily; they must obey the constraint.

So instead, derive the constraint algebraically: writing $x=b+c, y=c+a, z=a+b$ in HP means $\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$. Using $s=a+b+c$, we have $x=s-a,y=s-b,z=s-c$. Substituting and simplifying (cross multiplying, expanding $s^2=(a+b+c)^2$, and cancelling the $2ac-2ac$ terms) collapses neatly to:
$$a^2+c^2=2b^2$$

This is precisely the algebraic signature of three terms in Arithmetic Progression, since an AP requires the middle term to equal the average of its neighbours: $b^2=\dfrac{a^2+c^2}{2}$.

So whenever $(b+c),(c+a),(a+b)$ are in HP, it forces $a^2,b^2,c^2$ into AP. Final answer: (a) AP
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