Question

If $a,b$ and $c$ are in $A.P.,$ 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)

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