Question

If $a_1,a_2,...,a_n,...$ form a G.P and $a_i>0$ for all $i\ge 1$ 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 equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

we have

$a_{n+1}^2=a_n{a}_{n+2}\Rightarrow 2\log a_{n+1}=\log a_n+\log a_{n+2};2\log a_{n+4}=\log a_{n+3}+\log a_{n+5};2\log a_{n+7}=\log a_{n+6}+\log a_{n+8}$

substituting these values in second column of determinant, we gets

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{a}_n& \log a_n+\log a_{n+2}& \log a_{n+2}\\ \log a_{n+3}& \log a_{n+3}+\log a_{n+5}& \log a_{n+5}\\ \log a_{n+6}& \log a_{n+6}+\log a_{n+8}& \log a_{n+8}\end{array}\right|=\frac{1}{2}\left(0\right)=0(C_2\to C_2-C_1-C_3)$