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JEE Math Practice Question

If $$\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :

  1. A.220
  2. B.200
  3. C.240
  4. D.180

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