Question

If $a_1,a_2,a_3,a_4,a_5$ are consecutive terms of an arithmetic progression with common difference $3,$ then the value of $\left|\begin{array}{ccc}{a}_3^2& a_2& a_1\\ a_4^2& a_3& a_2\\ a_5^2& a_4& a_3\end{array}\right|$ is

• A. $0$
• B. $27$
• C. $81$
• D. $162$

Answer 1

Answer: (D)

Apply $R_3\leftrightarrow R_3-R_2,R_2\leftrightarrow R_2-R_1$
$\left|\begin{array}{ccc}{a}_3^2& a_2& a_1\\ a_4^2-a_3^2& a_3-a_2& a_2-a_1\\ a_5^2-a_4^2& a_4-a_3& a_3-a_2\end{array}\right|$
$=\left|\begin{array}{ccc}{a}_3^2& a_2& a_1\\ 3\left(a_3+a_4\right)& 3& 3\\ 3\left(a_4+a_5\right)& 3& 3\end{array}\right|$
Applying $C_3\leftrightarrow C_3-C_2$
$\left|\begin{array}{ccc}{a}_3^2& a_2& -3\\ 3\left(a_3+a_4\right)& 3& 0\\ 3\left(a_4+a_5\right)& 3& 0\end{array}\right|$
$=-3\left[9\left(a_3+a_4\right)-9\left(a_4+a_5\right)\right]$
$=-27\left[a_3-a_5\right]=-27\times -6=162$