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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