Question

If $a_1,a_2,....,a_n,...$ are in GP and $a_1>0$ for each $i$, then determinant $\Delta =\left|\begin{array}{ccc}\log a_n& \log a_{n+2}& \log a_{n+4}\\ \log a_{n+6}& \log a_{n+8}& \log a_{n+10}\\ \log a_{n+12}& \log a_{n+14}& \log a_{n+16}\end{array}\right|$ 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, $\Delta =\left|\begin{array}{ccc}\log a_n& \log a_{n+2}& \log a_{n+4}\\ \log a_{n+6}& \log a_{n+8}& \log a_{n+10}\\ \log a_{n+12}& \log a_{n+14}& \log a_{n+16}\end{array}\right|$
Applying $C_2\to 2C_2-C_1-C_3$
$=\left|\begin{array}{ccc}\log a_n& 2\log a_{n+2}-\log a_n-\log a_{n+4}& \log a_{n+4}\\ \log a_{n+6}& 2\log a_{n+8}-\log a_{n+6}-\log a_{n+10}& \log a_{n+10}\\ \log a_{n+12}& 2\log a_{n+14}-\log a_{n+12}-\log a_{n+16}& \log a_{n+16}\end{array}\right|$
$=\left|\begin{array}{ccc}\log a_n& 0& {\log }_{a_{n+4}}\\ \log a_{n+6}& 0& \log a_{n+10}\\ \log a_{n+12}& 0& \log a_{n+16}\end{array}\right|=0.$