JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 12)
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
$$4 \alpha+2 \beta$$
0
$$2 n$$
$$2 \alpha+4 \beta$$
Explanation
$$A_r=\left|\begin{array}{ccc} r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2} \end{array}\right|$$
$${A_r} = 2\left| {\matrix{ r & 1 & {{{{n^2}} \over 2} + \alpha } \cr {2r} & 2 & {{{{n^2}} \over 2} - \beta } \cr {3r - 2} & 3 & {{{n(3n - 1)} \over 2}} \cr } } \right|$$
$$R_1 \rightarrow R_1-R_2$$
$$ = 2\left| {\matrix{ 0 & 1 & {\alpha + {\beta \over 2}} \cr r & 2 & {{{{n^2}} \over 2} - \beta } \cr {3r - 2} & 3 & {{{n(3n - 1)} \over 2}} \cr } } \right|$$
$$\begin{aligned} & =2\left(\alpha+\frac{\beta}{2}\right)(3 r-3 r+2) \\ & A_r=4 \alpha+2 \beta \\ & 2 A_{10}-A_8=4 \alpha+2 \beta \end{aligned}$$
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