## Two cute problems in HP : IITJEE Foundations\Mains, pre RMO

Problem 1:

If $a^{2}, b^{2}, c^{2}$ are in AP, show that $b+c, c+a, a+b$ are in HP.

Proof 1:

Note that a straight forward proof is not so easy.

Below is a nice clever solution:

By adding $ab+bc+ca$ to each term, we see that:

$a^{2}+ab+ac+bc, b^{2}+ab+ac+bc, c^{2}+ab+ac+bc$ are in AP.

that is, $(a+b)(a+c), (b+c)(b+a), (c+a)(c+b)$ are in AP.

Dividing each term by $(a+b)(b+c)(c+a)$.

$\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP.

that is, $b+c, c+a, a+b$ are in HP.

QED.

Problem 2:

If the $p^{th}, q^{th}, r^{th}, s^{th}$ terms of an AP are in GP, show that $p-q, q-r, r-s$ are in GP.

Proof 2:

Once again a straight forward proof is not at all easy.

Below is a “bingo” sort of proof 🙂

With the usual notation, we have

$\frac{a+(p-1)d}{a+(q-1)d} = \frac{a+(q-1)d}{a+(r-1)d} = \frac{a+(r-1)d}{a+(s-1)d}$

Hence, each of the ratios is equal to

$\frac{(a+(p-1)d)-(a+(q-1)d)}{(a+(q-1)d)-(a+(r-1)d)} = \frac{(a+(q-1)d)-(a+(r-1)d)}{(a+(r-1)d)-(a+(s-1)d)}$

which in turn is equal  to $\frac{p-q}{q-r} = \frac{q-r}{r-s}$

Hence, $p-q, q-r, r-s$ are in GP.

Cheers,

Nalin Pithwa

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