Question

If $x, y, z$ are in A.P., then the value of the determinant
$\begin{vmatrix}a+2&a+3&a+2x\\ a+3&a+4&a+2y\\ a+4&a+5&a+2z\end{vmatrix}$ is

• A. 1
• B. 0
• C. 2a
• D. a

Answer 1

Answer: (B)

As $x, y, z$ are in A.P. Therefore, $x + z - 2y = 0$
Now, $\begin{vmatrix}a+2&a+3&a+2x\\ a+3&a+4&a+2y\\ a+4&a+5&a+2z\end{vmatrix}$
$\begin{vmatrix}0&0&2\left(x+z-2y\right)\\ a+3&a+4&a+2y\\ a+4&a+5&a+2z\end{vmatrix}$
[applying $\left.R_{1} \rightarrow R_{1}+R_{3}-2 R_{2}\right]$
$\begin{vmatrix}0&0&0\\ a+3&a+4&a+2y\\ a+4&a+5&a+2z\end{vmatrix}$
$[\because x+z-2 y=0]$
$=0$