If `t_1, t_2 ,t_3,t_4,t_5` be in A.P. common difference `d` then the value of D `=|(t_2_t_3,t_2,t_1),(t_3_t_4,t_3,t_2),(t_4_t_5,t_4,t_3)|` is

Options

A. `0`
B. `2d^2`
C. `2d^3`
D. `2d^4`

Answer

To solve the given problem, we need to evaluate the determinant \( D = \begin{vmatrix} t_2 - t_3 & t_2 & t_1 \\ t_3 - t_4 & t_3 & t_2 \\ t_4 - t_5 & t_4 & t_3 \end{vmatrix} \) given that \( t_1, t_2, t_3, t_4, t_5 \) are in arithmetic progression (A.P.) with a common difference \( d \).

### Step-by-Step Solution:

1. **Express the terms in terms of \( t_1 \) and \( d \)**:
- Let \( t_1 = a \).
- Then, \( t_2 = a + d \),
- \( t_3 = a + 2d \),
- \( t_4 = a + 3d \),
- \( t_5 = a + 4d \).

2. **Substitute these values into the determinant**:
\[ D = \begin{vmatrix} (a + d) - (a + 2d) & (a + d) & a \\ (a + 2d) - (a + 3d) & (a + 2d) & (a + d) \\ (a + 3d) - (a + 4d) & (a + 3d) & (a + 2d) \end{vmatrix} \]

3. **Simplify the determinant**:
- The first row becomes:
\[ (a + d) - (a + 2d) = -d, \quad (a + d), \quad a \]
- The second row becomes:
\[ (a + 2d) - (a + 3d) = -d, \quad (a + 2d), \quad (a + d) \]
- The third row becomes:
\[ (a + 3d) - (a + 4d) = -d, \quad (a + 3d), \quad (a + 2d) \]
- Thus, the determinant \( D \) can be rewritten as:
\[ D = \begin{vmatrix} -d & (a + d) & a \\ -d & (a + 2d) & (a + d) \\ -d & (a + 3d) & (a + 2d) \end{vmatrix} \]

4. **Factor out \(-d\) from each row**:
\[ D = -d \begin{vmatrix} 1 & \frac{(a + d)}{-d} & \frac{a}{-d} \\ 1 & \frac{(a + 2d)}{-d} & \frac{(a + d)}{-d} \\ 1 & \frac{(a + 3d)}{-d} & \frac{(a + 2d)}{-d} \end{vmatrix} \]
- This simplifies to:
\[ D = -d \cdot (-d) \begin{vmatrix} 1 & -\frac{(a + d)}{d} & -\frac{a}{d} \\ 1 & -\frac{(a + 2d)}{d} & -\frac{(a + d)}{d} \\ 1 & -\frac{(a + 3d)}{d} & -\frac{(a + 2d)}{d} \end{vmatrix} \]

5. **Evaluate the determinant**:
- Notice that the first column is identical in all rows, which implies that the determinant is zero:
\[ D = 0 \]

### Final Answer:
Thus, the value of the determinant \( D \) is \( 0 \).