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# Choose the correct answer. If a, b, c, are in A.P, then the determinant $\begin{vmatrix} x+2 & x+3 &x+2a \\ x+3 & x+4 &x+2b \\ x+4 & x+5 &x+2c \end{vmatrix}$ is:  $(A)\;0 \hspace{20 mm} (B)\;1 \hspace{20 mm} (C)\; x \hspace{20 mm} (D)\; 2x$

Toolbox:
• If each column of a row (or a column) of a determinant is multiplied by a constant ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any column.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\Delta=\begin{vmatrix}x+2 & x+3 & x+2a\\x+3 & x+4 & x+2b\\x+4 &x+5 & x+2c\end{vmatrix}$

Apply $R_1\rightarrow R_1+R_2$ and $R_2\rightarrow R_2-R_3$

$\Delta=\begin{vmatrix}-1 & -1 & 2a-2b\\-1 & -1 & 2b-2c\\x+4 &x+5 & x+2c\end{vmatrix}$

$\Delta=\begin{vmatrix}-1 & -1 & 2(a-b)\\-1 & -1 & 2(b-c)\\x+4 &x+5 & x+2c\end{vmatrix}$

Apply $C_1\rightarrow C_1+C_2$

$\Delta=\begin{vmatrix}0 & -1 & 2(a-b)\\0 & -1 & 2(b-c)\\x+4 &x+5 & x+2c\end{vmatrix}$

Now expanding along $C_1$ we get,

$\Delta=0-0+(x+4)[-1\times 2(a-b)+1\times 2(b-c)]$

$\Delta=(x+4)[-2a+b+2b-2c]$

But given a,b,c are in A.P;hence 2b=a+c.

Therefore $\Delta=(x+4)[-2a+b+a+c-2c]$

$\;\;\;\;\qquad\;\;\;\;\;=(x+4)[-a-c+b]$

$\;\;\;\;\qquad\;\;\;\;\;=(x+4)[-(a+c)+b]$ But (a+c=b)

Therefore $\Delta=(x+4)[-b+b]=0$

Hence the correct answer is A

edited Mar 5, 2013