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. $\frac{a+b}{2}$
• B. $ab$
• C. $0$
• D. $abc$

Answer 1

The correct option is C $0$

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

using $R_3=R_3-R_2\text{\&}{R}_2=R_2-R_1$

$\Delta =\left|\begin{array}{ccc}x+1& x+2& x+a\\ 1& 1& b-a\\ 1& 1& c-b\end{array}\right|$

Now since $a,,b,c\in$A.P. $\Rightarrow c-b=b-a$

Hence $\Delta =0$, since $R_3$ and $R_2$ are same.