A) 0
B) 1
C) 2
D) \[-2\]
Correct Answer: A
Solution :
\[\because \]\[{{a}_{1}},{{a}_{2}},....{{a}_{n}}\]are in GP. Then, \[{{a}_{n}}={{a}_{1}}{{r}^{n-1}}\] \[\Rightarrow \] \[\log {{a}_{n}}-\log {{a}_{1}}+(n-1)\log r\] \[{{a}_{n+1}}={{a}_{1}}{{r}^{n}}\] \[\Rightarrow \] \[\log {{a}_{n+1}}=\log {{a}_{1}}+n\log r\] \[{{a}_{n+2}}={{a}_{1}}{{r}^{n+1}}\] \[\Rightarrow \] \[\log {{a}_{n+2}}=\log {{a}_{1}}+(n+1)\log r\] Similarly, \[{{a}_{n+8}}={{a}_{1}}{{r}^{n+7}}\] \[\Rightarrow \] \[\log {{a}_{n+8}}=\log {{a}_{1}}+(n+7)\log r\] Now, \[\left| \begin{matrix} \log {{a}_{n}} & \log {{a}_{n+1}} & {{a}_{n+2}} \\ \log {{a}_{n+3}} & \log {{a}_{n+4}} & \log {{a}_{n+5}} \\ \log {{a}_{n+6}} & \log {{a}_{n+7}} & \log {{a}_{n+8}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \log {{a}_{1}}+(n-1)\log r & \log {{a}_{1}}+n\log r \\ \log {{a}_{1}}+(n+2)\log r & \log {{a}_{1}}+(n+6)\log r \\ \log {{a}_{1}}+(n+5)\log r & \log {{a}_{1}}+(n+6)\log r \\ \end{matrix} \right.\] \[\left. \begin{matrix} \log {{a}_{1}}+(n+1)\log r \\ \log {{a}_{1}}+(n+4)\log r \\ \log {{a}_{1}}+(n+7)\log r \\ \end{matrix} \right|\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\] \[=\left| \begin{matrix} \log {{a}_{1}}+(n+1)\log r & \log {{a}_{1}}+n\log r \\ 3\log r & 3\log r \\ 3\log r & 3\log r \\ \end{matrix} \right.\] \[\left. \begin{matrix} \log {{a}_{1}}+(n+1)\log r \\ 3\log r \\ 3\log r \\ \end{matrix} \right|\] \[=0\] [\[\because \]two rows are same]
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