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Question: If \(a_{1},a_{2},a_{3},.......,a_{n},......\) are in G.P. then the value of the determinant \(\left|...

If a1,a2,a3,.......,an,......a_{1},a_{2},a_{3},.......,a_{n},...... are in G.P. then the value of the determinant ∣log⁡anlog⁡an+1log⁡an+2log⁡an+3log⁡an+4log⁡an+5log⁡an+6log⁡an+7log⁡an+8∣\left| \begin{matrix} \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{matrix} \right| is

A

–2

B

1

C

2

D

0

Answer

0

Explanation

Solution

∵\because a1,a2,a3,.......,an,......a_{1},a_{2},a_{3},.......,a_{n},...... are in G.P.

∴an+12=an.an+2⇒2log⁡an+1=log⁡an+log⁡an+2\therefore a_{n + 1}^{2} = a_{n}.a_{n + 2} \Rightarrow 2\log a_{n + 1} = \log a_{n} + \log a_{n + 2}

an+42=an+3.an+5⇒2log⁡an+4=log⁡an+3+log⁡an+5a_{n + 4}^{2} = a_{n + 3}.a_{n + 5} \Rightarrow 2\log a_{n + 4} = \log a_{n + 3} + \log a_{n + 5}

an+72=an+6.an+8⇒2log⁡an+7=log⁡an+6+log⁡an+8a_{n + 7}^{2} = a_{n + 6}.a_{n + 8} \Rightarrow 2\log a_{n + 7} = \log a_{n + 6} + \log a_{n + 8}

Putting these values in the second column of the given

determinant, we get

\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{matrix} \right| = \frac{1}{2}(0) = 0$$ a [![](https://cdn.pureessence.tech/canvas_477.png?top_left_x=246&top_left_y=300&width=300&height=273)$c_{2}$is the sum of the elements, first identical with $c_{1}$ and second with $c_{3}$]