Mathematics · 2023
JEE Main · 31 January 2023, Shift 2 · Q65
Let a_1, a_2, a_3, … be an A.P. If a_7=3, the product a_1 a_4 is minimum and the sum of its first n terms is zero, then n!-4 a_n ( n +2) is equal to:
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots$ be an A.P. If $\displaystyle \mathrm{a}_7=3$, the product $\displaystyle \mathrm{a}_1 \mathrm{a}_4$ is minimum and the sum of its first n terms is zero, then $\displaystyle \mathrm{n}!-4 \mathrm{a}_{\mathrm{n}(\mathrm{n}+2)}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 24$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.