question_answer

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},............\]are positive numbers in G.P. then the value of \[\left| \begin{matrix}    \log {{a}_{n}} & \log {{a}_{n+1}} & \log {{a}_{n+2}}  \\    \log {{a}_{n+1}} & \log {{a}_{n+2}} & {{\operatorname{loga}}_{n+3}}  \\    \log {{a}_{n+2}} & \log {{a}_{n+3}} & \log {{a}_{n+4}}  \\ \end{matrix} \right|\]

A) \[1\]

B) \[4\]

C) \[3\]

D) \[0\]

Correct Answer: D

Solution :

[d] If the G.P be \[a,ar,a{{r}^{2}},....\] then \[{{a}_{n}}=a{{r}^{n-1}}\] \[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] gives,             \[=\left| \begin{matrix}    \log a+(n-1)log\,r & \log a+n\log r & \log a+(n+1)log\,r  \\    \log \,r & lor\,r & \log \,r  \\    \log \,r & \log \,r & \log \,r  \\ \end{matrix} \right|\]             = 0, since \[{{R}_{2}}\] and \[{{R}_{3}}\] are identical.