Question

If a, b, c are in A. P. then the value of

$$\begin{vmatrix} x+1 & x+2 & x+a \\[2pt] x+2 & x+3 & x+b \\[2pt] x+3 & x+4 & x+c \end{vmatrix}$$

is:

Answer 1

0

Answer 2

Step 1. Observe the first two columns differ by

$$\bigl[(x+2)-(x+1),\,(x+3)-(x+2),\,(x+4)-(x+3)\bigr]^T=[1,1,1]^T.$$

Perform column operation $C_2\to C_2-C_1$:

$$\begin{vmatrix} x+1 & 1 & x+a\\[2pt] x+2 & 1 & x+b\\[2pt] x+3 & 1 & x+c \end{vmatrix}.$$

Step 2. Expand along the 2nd column:

$$D=-\det\!\begin{pmatrix}x+2 & x+b\\ x+3 & x+c\end{pmatrix} +\det\!\begin{pmatrix}x+1 & x+a\\ x+3 & x+c\end{pmatrix} -\det\!\begin{pmatrix}x+1 & x+a\\ x+2 & x+b\end{pmatrix}.$$

Compute each minor using $b=a+d,\ c=a+2d$: All coefficients of $x$ cancel, and the constant term becomes $-c+2b-a$.
Since $b=a+d,\ c=a+2d$,

$$-c+2b-a=-(a+2d)+2(a+d)-a=0.$$

Hence $D=0$.