Question:
If $a, b, c$ are in A.P., then the determinant
$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$
(a) 0
(b) 1
(c) $x$
(d) $2 x$
Solution:
(a) 0
$\mid x+2 x+3 \quad x+2 a$
$x+3 x+4 \quad x+2 b$
$x+4 x+5 \quad x+2 c \mid$
$=\mid \begin{array}{lll}0 & 0 & 2(a+c-2 b)\end{array}$
$x+3 \quad x+4 \quad x+2 b$
$\begin{array}{llll}x+4 & x+5 & x+2 c \mid & {\left[\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-\mathrm{R}_{2}, \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}\right]}\end{array}$
$=\mid \begin{array}{lll}0 & 0 & 0\end{array}$
$\begin{array}{lll}x+3 & x+4 & x+2 b\end{array}$
$\begin{array}{ll}x+4 & x+5 & x+2 c\end{array}$ $[\because a, b, c$ are in A.P. $]$
$=0$