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If $a, b, c$ are in A.P., then the value of $\left|\begin{array}{ccc}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$ is
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The correct answer is:
$0$
$A=\left|\begin{array}{lll}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$
Applying $C_2 \rightarrow C_2-C_1$, we get,
$\Rightarrow A=\left|\begin{array}{lll}x+2 & 1 & x+a \\ x+4 & 1 & x+b \\ x+6 & 1 & x+c\end{array}\right|$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$
$\Rightarrow \quad A=\left|\begin{array}{ccc}x+2 & 1 & x+a \\ 2 & 0 & b-a \\ 4 & 0 & c-a\end{array}\right|=-1(2 c-2 a-4 b+4 a)$
$=2(2 b-c-a)$
$a, b, c$ are in A.P. $\Rightarrow A=0$.
Applying $C_2 \rightarrow C_2-C_1$, we get,
$\Rightarrow A=\left|\begin{array}{lll}x+2 & 1 & x+a \\ x+4 & 1 & x+b \\ x+6 & 1 & x+c\end{array}\right|$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$
$\Rightarrow \quad A=\left|\begin{array}{ccc}x+2 & 1 & x+a \\ 2 & 0 & b-a \\ 4 & 0 & c-a\end{array}\right|=-1(2 c-2 a-4 b+4 a)$
$=2(2 b-c-a)$
$a, b, c$ are in A.P. $\Rightarrow A=0$.
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