In an arithmetic sequence `a_(1),a_(2),a_(3), . . . . .,a_(n)`,
`Delta=|{:(a_(m),a_(n),a_(p)),(m,n,p),(1,1,1):}|` equals

Options

A. 1
B. -1
C. 0
D. mnp

Answer

To solve the problem, we need to find the value of the determinant \( \Delta = \begin{vmatrix} a_m & a_n & a_p \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \) where \( a_m, a_n, a_p \) are terms of an arithmetic sequence.

### Step-by-Step Solution:

1. **Identify the Terms of the Arithmetic Sequence**:
The \( m \)-th term of the arithmetic sequence can be expressed as:
\[ a_m = a + (m - 1)d \]
The \( n \)-th term is:
\[ a_n = a + (n - 1)d \]
The \( p \)-th term is:
\[ a_p = a + (p - 1)d \]

2. **Set Up the Determinant**:
The determinant we need to evaluate is:
\[ \Delta = \begin{vmatrix} a_m & a_n & a_p \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \]

3. **Substitute the Terms into the Determinant**:
Substitute \( a_m, a_n, a_p \) into the determinant:
\[ \Delta = \begin{vmatrix} a + (m - 1)d & a + (n - 1)d & a + (p - 1)d \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \]

4. **Apply Column Operations**:
We can simplify the determinant by performing column operations. Let's subtract the second column from the first and the third column from the second:
\[ \Delta = \begin{vmatrix} (a + (m - 1)d) - (a + (n - 1)d) & (a + (n - 1)d) - (a + (p - 1)d) & a + (p - 1)d \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} (m - n)d & (n - p)d & a + (p - 1)d \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \]

5. **Factor Out Common Terms**:
We can factor \( d \) out of the first row:
\[ \Delta = d \begin{vmatrix} m - n & n - p & a + (p - 1)d \\ m & n & p \\ 1 & 1 & 1 \end{vmatrix} \]

6. **Expand the Determinant**:
Now we can expand this determinant. However, notice that if we perform further operations, we will eventually see that the determinant will simplify to zero:
\[ \Delta = d \left( (m - n)(n - p) - (n - p)(m - n) \right) \]
This results in:
\[ \Delta = 0 \]

### Conclusion:
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]