Question

If $ {{a}_{1}},{{a}_{2}},....,{{a}_{n}},... $ are in GP and $ {{a}_{1}}>0 $ for each $i$, then determinant $ \Delta =\left| \begin{matrix} \log \,{{a}_{n}} & \log {{a}_{n+2}} & \log {{a}_{n+4}} \\ \log {{a}_{n+6}} & \log {{a}_{n+8}} & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & \log {{a}_{n+14}} & \log {{a}_{n+16}} \\ \end{matrix} \right| $ is equal to:

• A. 0
• B. 1
• C. 2
• D. $ n $

Answer 1

Answer: (A)

$ \because $ $ {{a}_{1}},{{a}_{2}},...,{{a}_{n}} $ are also in GP
$ \Rightarrow $ $ {{a}_{n}},{{a}_{n+2}},{{a}_{n+4}},.... $ are also GP
Now, $ {{({{a}_{n+2}})}^{2}},={{a}_{n}}.{{a}_{n+4}} $
$ 2\log ({{a}_{n+2}})=\log \,{{a}_{n}}+\log \,{{a}_{n+4}} $
Similarly $ 2\log ({{a}_{n+8}})=\log \,{{a}_{n+6}}+\log {{a}_{n+10}} $
Now, $ \Delta =\left| \begin{matrix} \log {{a}_{n}} & \log {{a}_{n+2}} & \log \,{{a}_{n+4}} \\ \log {{a}_{n+6}} & \log {{a}_{n+8}} & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & \log {{a}_{n+14}} & \log {{a}_{n+16}} \\ \end{matrix} \right| $
Applying $ {{C}_{2}}\to 2{{C}_{2}}-{{C}_{1}}-{{C}_{3}} $
$ =\left| \begin{matrix} \log {{a}_{n}} & 2\log {{a}_{n+2}}-\log {{a}_{n}}-\log {{a}_{n+4}} & \log {{a}_{n+4}} \\ \log {{a}_{n+6}} & 2\log {{a}_{n+8}}-\log {{a}_{n+6}}-\log {{a}_{n+10}} & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & 2\log {{a}_{n+14}}-\log {{a}_{n+12}}-\log {{a}_{n+16}} & \log \,{{a}_{n+16}} \\ \end{matrix} \right| $
$ =\left| \begin{matrix} \log {{a}_{n}} & 0 & {{\log }_{{{a}_{n+4}}}} \\ \log {{a}_{n+6}} & 0 & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & 0 & \log {{a}_{n+16}} \\ \end{matrix} \right|=0. $