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JEE Math Practice Question

For $$\alpha, \beta \in \mathbb{R}$$ and a natural number $$n$$, let $$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$. Then $$2 A_{10}-A_8$$ is

  1. A.$$4 \alpha+2 \beta$$
  2. B.0
  3. C.$$2 n$$
  4. D.$$2 \alpha+4 \beta$$

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