If a_{1}, a_{2}, a_{3}, ldots, a_{9} are in A | vedclass

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(D) Let the common difference of the $AP$ be $d$. Then $a_{n} = a_{1} + (n-1)d$.Applying the row operation $R_{1} \rightarrow R_{1} + R_{3} - 2R_{2}$:

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$0$

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(D) Let the common difference of the $AP$ be $d$. Then $a_{n} = a_{1} + (n-1)d$.Applying the row operation $R_{1} \rightarrow R_{1} + R_{3} - 2R_{2}$:The elements of the first row become:$a_{1} + a_{7} - 2a_{4} = (a_{1}) + (a_{1} + 6d) - 2(a_{1} + 3d) = 2a_{1} + 6d - 2a_{1} - 6d = 0$.$a_{2} + a_{8} - 2a_{5} = (a_{1} + d) + (a_{1} + 7d) - 2(a_{1} + 4d) = 2a_{1} + 8d - 2a_{1} - 8d = 0$.$a_{3} + a_{9} - 2a_{6} = (a_{1} + 2d) + (a_{1} + 8d) - 2(a_{1} + 5d) = 2a_{1} + 10d - 2a_{1} - 10d = 0$.Since all elements of the first row are $0$,the value of the determinant is $0$.

If $a_{1}, a_{2}, a_{3}, \ldots, a_{9}$ are in $AP$,then the value of $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is

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