question_answer

If \[{{a}_{1}},{{a}_{2}}...{{a}_{n}}....\]form a G.P. and \[{{a}_{i}}\]>0, for all \[i\ge 1\], then\[\left| \begin{matrix}    \log {{a}_{n}} & \log {{a}_{n+1}} & \log {{a}_{n+2}}  \\    \log {{a}_{n+3}} & \log {{a}_{n+4}} & \log {{a}_{n+5}}  \\    \log {{a}_{n+6}} & \log {{a}_{n+7}} & \log {{a}_{n+8}}  \\ \end{matrix} \right|\]is equal to

A)  0                    

B)  1

C)  2        

D)  3

Correct Answer: A

Solution :

[a] we have, \[{{a}_{n+1}}^{2}={{a}_{n}}{{a}_{n+2}}\] \[\Rightarrow 2\log {{a}_{n+1}}=\log {{a}_{n}}+\log {{a}_{n+2}}\] Similarly, \[2\log {{a}_{n+4}}=\log {{a}_{n+3}}+\log {{a}_{n+5}}\] \[2\log {{a}_{n+7}}=\log {{a}_{n+6}}+\log {{a}_{n+8}}\] Substituting these values in second column of determinant, we get