Question

If a, b and c are in AP, then the value of $\left|\begin{array}{ccc}x+2& x+3& x+a\\ x+4& x+5& x+b\\ x+6& x+7& x+c\end{array}\right|\begin{array}{}\\ \\ \end{array}$ is

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

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Step 1. Find the value of given determinant:

$\because a,b,c$ are in A.P, then

$2b=a+c$

$\Rightarrow$$b-a=c-b$ or $a-b=b-c$

$△=\left|\begin{array}{ccc}x+2& x+3& x+a\\ x+4& x+5& x+b\\ x+6& x+7& x+c\end{array}\right|\begin{array}{}\\ \\ \end{array}$

Step 2. On Performing $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$, we get

$△=\left|\begin{array}{ccc}-2& -2& a-b\\ -2& -2& b-c\\ x+6& x+7& x+c\end{array}\right|\begin{array}{}\\ \\ \end{array}$

$△=\left|\begin{array}{ccc}-2& -2& a-b\\ -2& -2& a-b\\ x+6& x+7& x+c\end{array}\right|\begin{array}{}\\ \\ \end{array}$ $\left[\because a-b=b-c\right]$

$\because$The two rows are same then the value of $△$ is $0$

$\mathbf{\therefore }\mathbf{△}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence , Option ‘A’ is Correct.