Least Positive Integer for Negative Term in Harmonic Progression

Given that $a_1, a_2, a_3, \dots$ are in harmonic progression (HP), their reciprocals form an arithmetic progression (AP).

Let $b_n = \frac{1}{a_n}$ be the $n$-th term of the corresponding AP, and let $d$ be its common difference.

We are given: $a_1 = 5 \implies b_1 = \frac{1}{5}$ $a_{20} = 25 \implies b_{20} = \frac{1}{25}$

Using the formula for the $n$-th term of an AP, $b_n = b_1 + (n-1)d$, we have: $b_{20} = b_1 + 19d$ $\frac{1}{25} = \frac{1}{5} + 19d$

Solving for $d$: $19d = \frac{1}{25} - \frac{1}{5} = -\frac{4}{25}$ $d = -\frac{4}{25 \times 19} = -\frac{4}{475}$

We are looking for the least positive integer $n$ such that $a_n < 0$. Since $a_n = \frac{1}{b_n}$, the condition $a_n < 0$ is equivalent to $b_n < 0$: $b_n < 0$ $b_1 + (n-1)d < 0$ $\frac{1}{5} + (n-1)\left(-\frac{4}{475}\right) < 0$

Rearranging the inequality: $\frac{1}{5} < (n-1)\left(\frac{4}{475}\right)$ $n - 1 > \frac{1}{5} \times \frac{475}{4}$ $n - 1 > \frac{95}{4}$ $n - 1 > 23.75$ $n > 24.75$

Since $n$ must be an integer, the least positive integer $n$ satisfying this inequality is $n = 25$.

Thus, the correct option is D.