Mathematics · 2025
JEE Main · 29 January 2025, Shift 2 · Q22
Let a_1, a_2, …, a_2024 be an Arithmetic Progression such that a_1+( a_5+ a_10+ a_15+…+ a_2020)+ a_2024=2233. Then a_1+ a_2+ a_3+…+ a_2024 is equal…
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{2024}$ be an Arithmetic Progression
such that $\displaystyle \mathrm{a}_1+\left(\mathrm{a}_5+\mathrm{a}_{10}+\mathrm{a}_{15}+\ldots+\mathrm{a}_{2020}\right)+\mathrm{a}_{2024}=2233$. Then $\displaystyle \mathrm{a}_1+\mathrm{a}_2+\mathrm{a}_3+\ldots+\mathrm{a}_{2024}$ is equal to
$\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
11132
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.