Question

If $ a,{{a}_{2}},{{a}_{3}},........{{a}_{n}},.... $ are in GP, then the value of the determinant $ \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. 0
• B. 1
• C. 2
• D. $ -2 $

Answer 1

Answer: (A)

Since $ {{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}} $ $ \Rightarrow $ $ \log {{a}_{n+2}}=\log {{a}_{1}}+(n+1)\log r $ ????.. ????. ????. $ {{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}} & \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| $ $ =\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+3)\log r \\ \log {{a}_{1}}+(n+5)\log r & \log {{a}_{1}}+(a+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| $ Now, $ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} $ and $ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} $ $ \Rightarrow $ $ \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 $ (since two rows are identical)