$If a_{1}, a_{2}, \dots . ., a_{n}, \dots . form a G.P. And a_{i} > 0, for all i \geq 1,$
$then |\begin{array}{lll} \log a_{n} \log a_{n + 1} \log a_{n + 2} \\ \log a_{n + 3} \log a_{n + 4} \log a_{n + 5} \\ \log a_{n + 6} \log a_{n + 7} \log a_{n + 8} \end{array}| is equal to$

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answer is A.

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Detailed Solution

$We have, \Rightarrow a_{n + 1}^{2} = a_{n} a_{n + 2}$
$\Rightarrow 2 \log a_{n + 1} = \log a_{n} + \log a_{n + 2}$
Similarly
$\begin{aligned} 2 \log a_{n + 4} = \log a_{n + 3} + \log a_{n + 5} \\ 2 \log a_{n + 7} = \log a_{n + 6} + \log a_{n + 8} \end{aligned}$
Substituting these values in second column of determinant, we get
$\begin{array}{l} Δ = \frac{1}{2} |\begin{array}{ccc} \log a_{n} & \log a_{n} + \log a_{n + 2} & \log a_{n + 2} \\ \log a_{n + 3} & \log a_{n + 3} + \log a_{n + 5} & \log a_{n + 5} \\ \log a_{n + 6} & \log a_{n + 6} + \log a_{n + 8} & \log a_{n + 8} \end{array}| \\ = \frac{1}{2} (0) = 0 using C_{2} \to C_{2} - C_{1} - C_{3}\} \end{array}$