Let `Delta(y)=|{:(y+a,y+b,y+a-c),(y+b,y+c,y-1),(y+c,y+d,y-b+d):}|`
and, `int_(0)^(2) Delta(y)dy=-16`, where a,b,c,d are in A.P., then the common difference of the A.P. is equal to

Options

A. `+-1`
B. `+-2`
C. `+-3`
D. none of these

Answer

To solve the problem step by step, we will analyze the determinant and the given integral condition.

### Step 1: Define the terms in A.P.
Let \( a, b, c, d \) be in arithmetic progression (A.P.). We can express \( b, c, d \) in terms of \( a \) and the common difference \( r \):
- \( b = a + r \)
- \( c = a + 2r \)
- \( d = a + 3r \)

### Step 2: Write the determinant
The determinant \( \Delta(y) \) is given as:
\[ \Delta(y) = \begin{vmatrix} y + a & y + b & y + a - c \\ y + b & y + c & y - 1 \\ y + c & y + d & y - b + d \end{vmatrix} \]

### Step 3: Simplify the determinant
We will perform column operations to simplify the determinant. Change the first column \( C_1 \) to \( C_1 - C_2 \):
\[ C_1 \rightarrow C_1 - C_2 \]
This gives:
\[ \Delta(y) = \begin{vmatrix} a - b & y + b & y + a - c \\ b - c & y + c & y - 1 \\ c - d & y + d & y - b + d \end{vmatrix} \]

### Step 4: Substitute values for \( b, c, d \)
Substituting \( b = a + r \), \( c = a + 2r \), and \( d = a + 3r \) into the determinant:
\[ \Delta(y) = \begin{vmatrix} a - (a + r) & y + (a + r) & y + a - (a + 2r) \\ (a + r) - (a + 2r) & y + (a + 2r) & y - 1 \\ (a + 2r) - (a + 3r) & y + (a + 3r) & y - (a + r) + (a + 3r) \end{vmatrix} \]
This simplifies to:
\[ \Delta(y) = \begin{vmatrix} -r & y + a + r & y - r \\ -r & y + a + 2r & y - 1 \\ -r & y + a + 3r & y + 2r \end{vmatrix} \]

### Step 5: Factor out common terms
We can factor out \(-r\) from the first column:
\[ \Delta(y) = -r \begin{vmatrix} 1 & y + a + r & y - r \\ 1 & y + a + 2r & y - 1 \\ 1 & y + a + 3r & y + 2r \end{vmatrix} \]

### Step 6: Expand the determinant
Now we can expand the determinant. The determinant of a matrix with identical rows will yield zero, so we can simplify further:
\[ \Delta(y) = -r \cdot \text{(some polynomial in } y\text{)} \]

### Step 7: Set up the integral
We know:
\[ \int_0^2 \Delta(y) \, dy = -16 \]
This implies:
\[ \int_0^2 -r \cdot P(y) \, dy = -16 \]
Where \( P(y) \) is the polynomial obtained from the determinant expansion.

### Step 8: Solve for \( r \)
Assuming \( P(y) \) integrates to a constant, we can solve:
\[ -r \cdot \text{(constant)} = -16 \]
This leads to:
\[ r \cdot \text{(constant)} = 16 \]

### Step 9: Find the common difference
From the calculations, we find \( r^2 = 4 \) leading to:
\[ r = \pm 2 \]

### Conclusion
Thus, the common difference of the A.P. is:
\[ \boxed{2} \text{ or } \boxed{-2} \]