Mathematics · 2023
JEE Main · 6 April 2023, Shift 2 · Q10
If gcd ( m, n )=1 and 1^2-2^2+3^2-4^2+…..+(2021)^2-(2022)^2+(2023)^2=1012 m^2 n then m^2-n^2 is equal to
If $\displaystyle \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$ and
$$1^2-2^2+3^2-4^2+\ldots . .+(2021)^2-(2022)^2+(2023)^2=1012 m^2 n
$$
then $\displaystyle m^2-n^2$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 240$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.