1.For α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R and a natural number nnn, let An=∣n2r1+α2r2n2−β3r−23n(3n−1)2∣A_n = \begin{vmatrix} n^2 & r & 1+\alpha \\ 2r & 2 & n^2-\beta \\ 3r-2 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}An=n22r3r−2r231+αn2−β2n(3n−1). Then 2A10−A92A_{10} - A_92A10−A9 isa.4α+2β4\alpha + 2\beta4α+2βb.2α+4β2\alpha + 4\beta2α+4βc.2n2n2nd.000Login to continueOnly logged in users canattempt or see the solution.
