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# If $a_1,a_2,a_3.......a_n$ are in G.P then the value of the determinant $\begin{vmatrix}\log a_n&\log a_{n+1}&\log a_{n+2}\\\log a_{n+3}&\log a_{n+4}&\log a_{n+5}\\\log a_{n+6}&\log a_{n+7}&\log a_{n+8}\end{vmatrix}$ is

$(a)\;-2\qquad(b)\;1\qquad(c)\;2\qquad(d)\;0$

Let $r$ be the common ratio,then
$\begin{vmatrix}\log a_n&\log a_{n+1}&\log a_{n+2}\\\log a_{n+3}&\log a_{n+4}&\log a_{n+5}\\\log a_{n+6}&\log a_{n+7}&\log a_{n+8}\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\log a_1r^{n-1}&\log a_1r^n&\log a_1r^{n+1}\\\log a_1r^{n+2}&\log a_1r^{n+3}&\log a_1r^{n+4}\\\log a_1r^{n+5}&\log a_1r^{n+6}&\log a_1r^{n+7}\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\log a_1+(n-1)\log r&\log a_1+n\log r&\log a_1+(n+1)\log r\\\log a_1+(n+2)\log r&\log a_1+(n+3)\log r&\log a_2+(n+4)\log r\\\log a_1+(n+5)\log r&\log a_1+(n+6)\log r&\log a_2+(n+7)\log r\end{vmatrix}$
By applying $C_2\rightarrow C_2-\large\frac{1}{2}$$C_1-\large\frac{1}{2}$$C_3$
$\Rightarrow 0$
Hence (d) is the correct answer.