If `|(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|` = 0 then a,b,c are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None

Answer

To solve the problem, we need to evaluate the determinant of the given matrix and set it equal to zero. The matrix is:

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

### Step 1: Write down the determinant
The determinant of a 3x3 matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we can denote:

- \( a = x + 1 \)
- \( b = x + 2 \)
- \( c = x + a \)
- \( d = x + 2 \)
- \( e = x + 3 \)
- \( f = x + b \)
- \( g = x + 3 \)
- \( h = x + 4 \)
- \( i = x + c \)

### Step 2: Simplify the determinant
To simplify the calculation, we can perform row operations. Let's perform the operation \( R_2 \leftarrow 2R_2 - R_1 - R_3 \):

\[ R_2 = 2(x + 2, x + 3, x + b) - (x + 1, x + 2, x + a) - (x + 3, x + 4, x + c) \]

This results in:

\[ R_2 = (2x + 4 - (x + 1) - (x + 3), 2x + 6 - (x + 2) - (x + 4), 2b - a - c) \]

Calculating each term:

1. First term: \( 2x + 4 - x - 1 - x - 3 = 0 \)
2. Second term: \( 2x + 6 - x - 2 - x - 4 = 0 \)
3. Third term remains \( 2b - a - c \)

Thus, we have:

\[ \begin{vmatrix} x + 1 & x + 2 & x + a \\ 0 & 0 & 2b - a - c \\ x + 3 & x + 4 & x + c \end{vmatrix} \]

### Step 3: Expand the determinant
Now, we can expand the determinant along the second row:

\[ \text{det}(A) = 0 \cdot \text{det} + 0 \cdot \text{det} + (2b - a - c) \cdot \text{det} \]

The determinant simplifies to:

\[ (2b - a - c) \cdot \begin{vmatrix} x + 1 & x + a \\ x + 3 & x + c \end{vmatrix} \]

### Step 4: Set the determinant to zero
Since we know that the determinant equals zero, we have:

\[ 2b - a - c = 0 \]

This implies:

\[ 2b = a + c \]

### Step 5: Analyze the relationship
The equation \( 2b = a + c \) indicates that \( a, b, c \) are in Arithmetic Progression (AP).

### Final Conclusion
Thus, the values \( a, b, c \) are in **Arithmetic Progression (AP)**.

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