Let `a_(1),a_(2),a_(3), . . .` be a harmonic progression with `a_(1)=5anda_(20)=25`. The least positive integer n for which `a_(n)lt0`, is

Options

A. 22
B. 23
C. 24
D. 25

Explanation

To solve the problem, we need to find the least positive integer \( n \) for which \( a_n < 0 \) in a harmonic progression (HP) defined by \( a_1 = 5 \) and \( a_{20} = 25 \).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
In a harmonic progression, the reciprocals of the terms form an arithmetic progression (AP). Therefore, we can express the terms of the HP as:
\[ a_n = \frac{1}{b_n} \]
where \( b_n \) is an arithmetic progression.

2. **Finding the First Term and the 20th Term**:
Given:
\[ a_1 = 5 \quad \text{and} \quad a_{20} = 25 \]
The reciprocals are:
\[ b_1 = \frac{1}{a_1} = \frac{1}{5} \quad \text{and} \quad b_{20} = \frac{1}{a_{20}} = \frac{1}{25} \]

3. **Finding the Common Difference**:
Since \( b_n \) is an arithmetic progression, we can write:
\[ b_n = b_1 + (n-1)d \]
For \( n = 20 \):
\[ b_{20} = b_1 + 19d \]
Substituting the values:
\[ \frac{1}{25} = \frac{1}{5} + 19d \]
Rearranging gives:
\[ 19d = \frac{1}{25} - \frac{1}{5} \]
Finding a common denominator (which is 25):
\[ 19d = \frac{1 - 5}{25} = \frac{-4}{25} \]
Thus:
\[ d = \frac{-4}{25 \times 19} = \frac{-4}{475} \]

4. **Finding the General Term**:
The general term for \( b_n \) is:
\[ b_n = \frac{1}{5} + (n-1) \left(\frac{-4}{475}\right) \]
Simplifying:
\[ b_n = \frac{1}{5} - \frac{4(n-1)}{475} \]

5. **Finding \( n \) such that \( a_n < 0 \)**:
We need \( a_n < 0 \), which means \( b_n > 0 \):
\[ b_n = \frac{1}{5} - \frac{4(n-1)}{475} > 0 \]
Rearranging gives:
\[ \frac{1}{5} > \frac{4(n-1)}{475} \]
Multiplying both sides by 475:
\[ \frac{475}{5} > 4(n-1) \]
Simplifying:
\[ 95 > 4(n-1) \]
Dividing by 4:
\[ \frac{95}{4} > n - 1 \]
Therefore:
\[ n < \frac{95}{4} + 1 = 23.75 + 1 = 24.75 \]
Since \( n \) must be a positive integer, the largest integer satisfying this is \( n = 24 \).

6. **Finding the Least Positive Integer \( n \) for \( a_n < 0 \)**:
The least positive integer \( n \) for which \( a_n < 0 \) is:
\[ n = 25 \]

### Final Answer:
The least positive integer \( n \) for which \( a_n < 0 \) is \( \boxed{25} \).