Question

If $a,b,c$ are in $A.P.$ then $\left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $0$

Since a,b,c are in A.P

So, $2b=a+c\left(1\right)$

$\left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|$

$R_1\to R_1+R_3-2R_2$

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

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

$=$$0$