JEE Main 2025 Jan 29 Shift 2, Mathematics Q23: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Jan 29, Shift 2Mathematics

Let $a_{1}, a_{2}, \dots, a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + \dots + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + \dots + a_{2024}$ is equal to ________.

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Solution

The inner sum has indices $5, 10, 15, \dots, 2020$ , totalling $\frac{2020}{5} = 404$ terms. Pair them up around the centre:

$(a_{5} + a_{2020}), (a_{10} + a_{2015}), \dots$ — that is $202$ pairs, and in an A.P. each such pair equals $a_{1} + a_{2024}$ .

Including the standalone $a_{1} + a_{2024}$ , the given expression becomes $203 (a_{1} + a_{2024}) = 2233$ , so $a_{1} + a_{2024} = 11$ .

$S_{2024} = \frac{2024}{2} (a_{1} + a_{2024}) = 1012 \times 11 = 11132$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q23: Arithmetic, Geometric And Harmonic Progressions

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