If a, b, c are in A.P.,then the determinant l | vedclass

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(D) Let $\Delta = \left|\begin{array}{lll}x+2 & x+3 & x+2a \ x+3 & x+4 & x+2b \ x+4 & x+5 & x+2c\end{array}ight|$.Since $a, b, c$ are in $A.P.$,we

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$0$

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(D) Let $\Delta = \left|\begin{array}{lll}x+2 & x+3 & x+2a \ x+3 & x+4 & x+2b \ x+4 & x+5 & x+2c\end{array}ight|$.Since $a, b, c$ are in $A.P.$,we have $2b = a+c$.Substituting $2b = a+c$ in the determinant,we get:$\Delta = \left|\begin{array}{lll}x+2 & x+3 & x+2a \ x+3 & x+4 & x+a+c \ x+4 & x+5 & x+2c\end{array}ight|$.Applying row operations $R_1 ightarrow R_1 - R_2$ and $R_3 ightarrow R_3 - R_2$:$\Delta = \left|\begin{array}{ccc}-1 & -1 & a-c \ x+3 & x+4 & x+a+c \ 1 & 1 & c-a\end{array}ight|$.Applying $R_1 ightarrow R_1 + R_3$:$\Delta = \left|\begin{array}{ccc}0 & 0 & 0 \ x+3 & x+4 & x+a+c \ 1 & 1 & c-a\end{array}ight|$.Since all elements of the first row are $0$,the value of the determinant is $0$.Thus,the correct option is $D$.

If $a, b, c$ are in $A.P.$,then the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c\end{array}\right|$ is

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