JEE Main 2026 Jan 22 Shift 2, Mathematics Q24: Arithmetic, Geometric And Harmonic Progressions

JEE Main2026Jan 22, Shift 2Mathematics

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equal to ______.

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Solution

Let $a = b - d$ , $c = b + d$ with $d > 0$ (since $a < b < c$ ). From $a + b + c = 3 b = 1$ , $b = \frac{1}{3}$ .

From the GP condition: $(2 b^{2})^{2} = a^{2} c^{2} \Rightarrow 4 b^{4} = (a c)^{2} = (b^{2} - d^{2})^{2}$ .

Taking positive root: $2 b^{2} = b^{2} - d^{2}$ would give $d^{2} < 0$ (impossible), so $2 b^{2} = d^{2} - b^{2} \Rightarrow d^{2} = 3 b^{2} = \frac{1}{3}$ , $d = \frac{1}{3}$ .

$9 (a^{2} + b^{2} + c^{2}) = 9 [(b - d)^{2} + b^{2} + (b + d)^{2}] = 9 (3 b^{2} + 2 d^{2}) = 9 (\frac{1}{3} + \frac{2}{3}) = 9$ .

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JEE Main 2026 Jan 22 Shift 2, Mathematics Q24: Arithmetic, Geometric And Harmonic Progressions

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