Question

If $a_{n}(>0)$ be the $n^{\text {th }}$ term of a G.P. then
$\begin{vmatrix}\log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{vmatrix}$ is equal to

• A. 1
• B. 2
• C. -2
• D. 0

Answer 1

Answer: (D)

$\therefore \begin{vmatrix}\log a+(n-1) \log r & \log a+n \log r & \log a+(n+1) \log r \\ \log a+(n+2) \log r & \log a+(n+3) \log r & \log a+(n+4) \log r \\ \log a+(n+5) \log r & \log a+(n+6) \log r & \log a+(n+7) \log r\end{vmatrix}$
$R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$
$\therefore \begin{vmatrix}\log a+(n-1) \log r & \log a+n \log r & \log a+(n+1) \log r \\ 3 \log r & 3 \log r & 3 \log r \\ 6 \log r & 6 \log r & 6 \log r\end{vmatrix}$
$=0$