Question

If $a_1,a_2,....,a_n$, ..... are in G.P. then $\left|\begin{array}{ccc}\log a_n& \log a_{n+1}& \log a_{n+2}\\ \log a_{n+3}& \log a_{n+4}& \log a_{n+5}\\ \log a_{n+6}& \log a_{n+7}& \log a_{n+8}\end{array}\right|$ is

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

We know that $\frac{a_2}{a_1}=\frac{a_3}{a_2}=.......\frac{a_n}{a_{n-1}}=r$

which means that $a_n,a_{n+1}$ and $a_{n+2}$ are in G.P

So,

$a_{n+1}^2=a_n\times a_{n+2}$

Taking log on both the sides, we get

$2log{a}_{n+1}=log{a}_n+log{a}_{n+2}$

$2log{a}_{n+1}-log{a}_n-log{a}_{n+2}=0$ --------(1)

Similarly,

$2log{a}_{n+4}-log{a}_{n+3}-log{a}_{n-5}=0$ -------(2)

$2log{a}_{n+7}-log{a}_{n+6}-log{a}_{n+8}=0$ ---------(3)

Apply the operation on column, we get

$C_1\to 2C_2-C_1-C_3$

$det\left(A\right)=0$