Q110 / 180 · COMEDK 2015 · Mathematics
If $a, b, c$ are in $\mathrm{AP}$, then the value of the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is
- Correct answerA0
- B1
- C$x$
- D$2 x$
Solution
Let $\Delta=\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$
Applying $R_{2} \rightarrow 2 R_{2}$, we get
$$
\Delta=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
2 x+6 & 2 x+8 & 2 x+4 b \\
x+4 & x+5 & x+2 c
\end{array}\right|
$$
Applying $R_{2} \rightarrow R_{2}-\left(R_{1}+R_{3}\right)$, we get
$$
\Delta=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
0 & 0 & 4 b-(2 a+2 c) \\
x+4 & x+5 & x+2 c
\end{array}\right|
$$
$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{ccc}x+2 & x+3 & x+2 a \\ 0 & 0 & 0 \\ x+4 & x+5 & x+2 c\end{array}\right|\left[\begin{array}{c}\because a, b, c \text { are in AP } \\ \therefore 2 b=a+c\end{array}\right]$
$\Rightarrow \Delta=0$
Hence, the correct answer is option: 'A'.