Mathematics

Mathematics

Problem

23. If the first and the nth n^{\text {th }} term of a G.P. are aa and bb, respectively, and if P is the product of nn terms, prove that P2=(ab)n\mathrm{P}^{2}=(a b)^{n}.
Intuitive explanation

In a geometric progression, the terms are symmetrically arranged around the middle—the first term pairs with the last, the second with the second-to-last, and so on. The key insight is that each such pair multiplies to the same value: a⋅ba \cdot b. Since there are nn terms total, the product of all nn terms can be regrouped as nn copies of this pairing, giving you P=(ab)n/2P = (ab)^{n/2}, which immediately yields P2=(ab)nP^2 = (ab)^n.

Intuitive explanation

In a geometric progression, the terms are symmetrically arranged around the middle—the first term pairs with the last, the second with the second-to-last, and so on. The key insight is that each such pair multiplies to the same value: a⋅ba \cdot b. Since there are nn terms total, the product of all nn terms can be regrouped as nn copies of this pairing, giving you P=(ab)n/2P = (ab)^{n/2}, which immediately yields P2=(ab)nP^2 = (ab)^n.

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