Question:
Show that $\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$ where $a, b, c$ are in A.P.
Solution:
Given: $a, b, c$ are in A.P.
$2 b=a+c$
$\Delta=\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|$ [Applying $R_{2}=2 R_{2}$ ]
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}x+1 & x+2 & x+a \\ 2 x+4 & 2 x+6 & 2 x+2 b \\ x+3 & x+4 & x+c\end{array}\right|$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}x+1 & x+2 & x+a \\ 0 & 0 & 0 \\ x+3 & x+4 & x+c\end{array}\right|$ $[\because 2 b=a+c] \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-\left(R_{1}+R_{3}\right)\right]$
$\Delta=0$