For $\alpha ,\beta \in \mathrm{ℝ}$ and a natural number $n$, let ${A}_{r}=\left|\begin{matrix}r & 1 & \frac{{n}^{2}}{2}+\alpha \ 2r & 2 & {n}^{2}-\beta \ 3r-2 & 3 & \frac{n(3n-1)}{2}\end{matrix}\right|$. Then $2{A}_{10}-{A}_{8}$ is [JEE Main 2024, 6 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $4\alpha +2\beta$. $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \ 2 r & 2 & n^2-\beta \ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$ $2 A_{10}-A_8=\left|\begin{array}{ccc}20 & 1 & \frac{n^2}{2}+a \ 40 & 2 & n^2-\beta \ 56 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{n^2}{2}+a \ 16 & 2 & n^2-\beta \ 22 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$ $=\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+a \ 24 & 2 & n^2-\beta \ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$ $=\left|\begin{array}{ccc}0 & 1 & \frac{n^2}{2}+a \ 0 & 2 & n^2-\beta \ -2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$ $=-2\left(\left(n^2-\beta\right)-\left(n^2+2 \alpha\right)\right)$ $=-2(-\beta-2 \alpha)$ $=4 \alpha+2 \beta$

For \(\alpha ,\beta \in \mathrm{ℝ}\) and a natural number \(n\), let \({A}_{r}=\left|\begin{matrix}r & 1 & \frac{{n}^{2}…

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