Let $a_1,a_2,.....$ be in H.P. with $a_1=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_n<0$ is

$\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\dots$ are in A.P.
Let $d$be the common difference of this A.P., then
       $\frac{1}{a_{20}}-\frac{1}{a_1}=19d$
$\Rightarrow \frac{1}{25}-\frac{1}{5}=19d\Rightarrow d=-\frac{4}{475}$
Now,    $a_n<0\Rightarrow \frac{1}{5}-\frac{(n-1)4}{475}<0$
$\Rightarrow n-1>95/4\Rightarrow n\ge 25.$
Thus, least value of $n$is $25.$