A) 0
B) 1
C) 2
D) \[n\]
Correct Answer: A
Solution :
\[\because \]\[{{a}_{1}},{{a}_{2}},...,{{a}_{n}}\]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 \[{{c}_{2}}\to 2{{C}_{2}}-{{C}_{1}}-{{C}_{3}}\] \[=\left| \begin{matrix} \log {{a}_{n}} & 2\log {{a}_{n+2}}-\log {{a}_{n}}-\log {{a}_{n+4}} & \log {{a}_{n+4}} \\ \log {{a}_{n+6}} & 2\log {{a}_{n+8}}-\log {{a}_{n+6}}-\log {{a}_{n+10}} & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & 2\log {{a}_{n+14}}-\log {{a}_{n+12}}-\log {{a}_{n+16}} & \log \,{{a}_{n+16}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \log {{a}_{n}} & 0 & {{\log }_{{{a}_{n+4}}}} \\ \log {{a}_{n+6}} & 0 & \log {{a}_{n+10}} \\ \log {{a}_{n+12}} & 0 & \log {{a}_{n+16}} \\ \end{matrix} \right|=0.\]
You need to login to perform this action.
You will be redirected in
3 sec
