Question

If a, b, 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) zero
b) 1
c) x
c) 2x

Answer 1

Let $A=\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}x+2& x+3& x+2a\\ 0& 0& 1(2b-a-c)\\ x+4& x+5& x+2c\end{array}\right|$
(using $R_2\to 2R_2-R_1-R_3$)
But a,b,c are in AP. Using 2b=a+c, we get
$A=\frac{1}{2}\left|\begin{array}{ccc}x+2& x+3& x+2a\\ 0& 0& 0\\ x+4& x+5& x+2c\end{array}\right|$
[Since, all elements of $R_2$ are zero]
Hence, the correct option is (a).