If the pth term of the series `25, 22 (3)/(4), 20 (1)/(2), 18 (1)/(4),…..` is numerically the smallest, then p =

Options

A. 11
B. 12
C. 13
D. 14

Answer

To solve the problem, we need to determine the value of \( p \) such that the \( p \)-th term of the series \( 25, 22 \frac{3}{4}, 20 \frac{1}{2}, 18 \frac{1}{4}, \ldots \) is numerically the smallest.

### Step-by-Step Solution:

1. **Identify the first term and common difference:**
- The first term \( a = 25 \).
- To find the common difference \( d \), we calculate:
\[ d = 22 \frac{3}{4} - 25 = \frac{91}{4} - \frac{100}{4} = -\frac{9}{4} \]

2. **General formula for the \( n \)-th term of an arithmetic progression (AP):**
- The \( n \)-th term \( T_n \) can be expressed as:
\[ T_n = a + (n - 1) \cdot d \]
- Substituting \( a \) and \( d \):
\[ T_n = 25 + (n - 1) \cdot \left(-\frac{9}{4}\right) \]

3. **Set up the inequality for the \( p \)-th term to be non-negative:**
- We want \( T_p \geq 0 \):
\[ T_p = 25 + (p - 1) \cdot \left(-\frac{9}{4}\right) \geq 0 \]
- Rearranging gives:
\[ 25 - \frac{9}{4}(p - 1) \geq 0 \]

4. **Solve the inequality:**
- Rearranging the inequality:
\[ 25 \geq \frac{9}{4}(p - 1) \]
- Multiply both sides by \( 4 \):
\[ 100 \geq 9(p - 1) \]
- Expanding gives:
\[ 100 \geq 9p - 9 \]
- Rearranging:
\[ 9p \leq 109 \implies p \leq \frac{109}{9} \approx 12.11 \]

5. **Determine the largest integer \( p \):**
- Since \( p \) must be an integer, the largest possible value is \( p = 12 \).

6. **Verify if \( p = 12 \) gives the smallest term:**
- Calculate \( T_{12} \):
\[ T_{12} = 25 + (12 - 1) \cdot \left(-\frac{9}{4}\right) = 25 - \frac{99}{4} = 25 - 24.75 = 0.25 \]
- Check \( T_{13} \):
\[ T_{13} = 25 + (13 - 1) \cdot \left(-\frac{9}{4}\right) = 25 - \frac{108}{4} = 25 - 27 = -2 \]
- Since \( T_{12} = 0.25 \) and \( T_{13} = -2 \), the \( 12 \)-th term is indeed the smallest non-negative term.

### Conclusion:
Thus, the value of \( p \) such that the \( p \)-th term is numerically the smallest is:
\[ \boxed{12} \]