Question

Give that a, b, c are in AP. The determinant
$\left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|$ in its simpliest form is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|Operate{C}_1\to C_1-C_2=\left|\begin{array}{ccc}-1& x+2& x+a\\ -1& x+3& x+b\\ -1& x+4& x+c\end{array}\right|=(-1)\left|\begin{array}{ccc}1& x+2& x+a\\ 1& x+3& x+b\\ 1& x+4& x+c\end{array}\right|Operate{R}_1-R_2-R_1,R_3\to R_3-R_1=(-1)\left|\begin{array}{ccc}1& x+2& x+a\\ 0& 1& b-a\\ 0& 2& c-a\end{array}\right|=(-1)[c+a-2b]=0$
$[\because a,b,careinAP\Rightarrow 2b=a+c]$