JEE Main 2026 Jan 28 Shift 1, Mathematics Q21: Arithmetic, Geometric And Harmonic Progressions

JEE Main2026Jan 28, Shift 1Mathematics

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ____.

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Solution

Let the first three G.P. terms be $\frac{A}{r}, A, A r$ . Product: $A^{3} = 27 \Rightarrow A = 3$ .

Sum: $S = 3 (\frac{1}{r} + 1 + r) = 3 + 3 (r + \frac{1}{r})$ .

For real nonzero $r$ : $r + \frac{1}{r} \geq 2$ (if $r > 0$ ) or $r + \frac{1}{r} \leq - 2$ (if $r < 0$ ).

So the attainable range of $S$ is $S \leq 3 + 3 (- 2) = - 3$ or $S \geq 3 + 3 (2) = 9$ , i.e. $S \in R ∖ (- 3, 9)$ .

Thus $(a, b) = (- 3, 9)$ and $a^{2} + b^{2} = 9 + 81 = 90$ .

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JEE Main 2026 Jan 28 Shift 1, Mathematics Q21: Arithmetic, Geometric And Harmonic Progressions

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