Question

If $a, b$ and $c$ are in A.P., then the value of $\begin{vmatrix} {x+2}&{x+3} &{x+a}\\ {x+4}&{x+5}& {x+b} \\ {x+6}&{x+7} &{x+c}\\ \end{vmatrix}$ is

• A. $x-(a + b + c)$
• B. $9x^2 + a + b + c$
• C. $0$
• D. $a + b + c$

Answer 1

Answer: (C)

Let $\Delta= \begin{vmatrix}x+2 & x+3 & x+ a \\ x+4 & x+5 & x +b \\ x+6 & x+7 & x +c\end{vmatrix}$
Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$, we get
$\Delta= \begin{vmatrix} -2 & -2 & a-b \\ -2 & -2 & b-c \\ x+6 & x+7 & x +c \end{vmatrix}$
Since, $a, b$ and $c$ are in $AP$.
$\therefore a-b=b-c$
Thus, Rows $R_{1}$ and $R_{2}$ are identical.
Hence, $\Delta=0$