Question

If $a,b$ and $c$ are in $AP$, then determinant $\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|$ is

• A. $0$
• B. $1$
• C. $x$
• D. $2x$

Answer 1

Answer: (A)

$0$

Let $△=\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}x+2& x+3& x+2a\\ 0& 0& 2(2b-a-c)\\ x+4& x+5& x+2c\end{array}\right|$ ...... (using $R_2\to 2R_2-R_1-R_2)$

But $a,b$ and $c$ are in $AP$ using $2b=a+c$, we get
$△=\frac{1}{2}\left|\begin{array}{ccc}x+2& x+3& x+2a\\ 0& 0& 0\\ x+4& x+5& x+2c\end{array}\right|$

$=0$ ........ (Since, all elements of $R_2$ are zero.)

Hence, $\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|=0$