Let \(a_{1}, a_{2}, a_{3}, \ldots . .\) be in harmonic progression with \(a_{1}=5\) and \(a_{20}=25\). The least positive integer \(n\) for which \(a_{n} < 0\) is

\(\because a_{1}, a_{2}, a_{3}, \ldots \ldots\) are in H.P.
\(\therefore \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \ldots \ldots\) are in A.P.
\(\therefore \frac{1}{a_{1}}=\frac{1}{5} \text { and } \frac{1}{a_{20}}=\frac{1}{25} \)
\(\frac{1}{a_{1}}+19 d=\frac{1}{a_{20}} \Rightarrow \frac{1}{5}+19 d=\frac{1}{25} \Rightarrow d=\frac{-4}{475} \)
Now \(\frac{1}{a_{n}}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right) \)
Clearly \(a_{n} < 0, \text { if } \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5}-\frac{4 n}{475}+\frac{4}{475} < 0 \)
\(\Rightarrow -4 n < -99 \text { or } n>\frac{99}{4}=24 \frac{3}{4} \therefore n \geq 25 \)
\(\therefore\) Least value of \(n \text { is } 25.\)