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 $\begin{vmatrix} a_{3}^{2} & a_{2} & a_{1} \\ a_{4}^{2} & a_{3} & a_{2} \\ a_{5}^{2} & a_{4} & a_{3} \end{vmatrix}$ is

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

Answer 1

Answer: (D)

Apply $R_{3}\leftrightarrow R_{3}-R_{2},R_{2}\leftrightarrow R_{2}-R_{1}$
$\begin{vmatrix} a_{3}^{2} & a_{2} & a_{1} \\ a_{4}^{2}-a_{3}^{2} & a_{3}-a_{2} & a_{2}-a_{1} \\ a_{5}^{2}-a_{4}^{2} & a_{4}-a_{3} & a_{3}-a_{2} \end{vmatrix}$
$=\begin{vmatrix} a_{3}^{2} & a_{2} & a_{1} \\ 3\left(a_{3} + a_{4}\right) & 3 & 3 \\ 3\left(a_{4} + a_{5}\right) & 3 & 3 \end{vmatrix}$
Applying $C_{3}\leftrightarrow C_{3}-C_{2}$
$\begin{vmatrix} a_{3}^{2} & a_{2} & -3 \\ 3\left(a_{3} + a_{4}\right) & 3 & 0 \\ 3\left(a_{4} + a_{5}\right) & 3 & 0 \end{vmatrix}$
$=-3\left[9 \left(a_{3} + a_{4}\right) - 9 \left(a_{4} + a_{5}\right)\right]$
$=-27\left[a_{3} - a_{5}\right]=-27\times -6=162$