Mathematics - Sequence & Series Question with Solution | TestHub

$\because a_1,a_2,a_3,\dots \dots$ are in H.P. $\begin{array}{l}\therefore \frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}\dots \dots \text{ are in A.P. }\\ \therefore \frac{1}{a_1}=\frac{1}{5}\text{ and }\frac{1}{a_{20}}=\frac{1}{25}\\ \frac{1}{a_1}+19d=\frac{1}{a_{20}}\Rightarrow \frac{1}{5}+19d=\frac{1}{25}\Rightarrow d=\frac{-4}{475}\\ \text{ Now }\frac{1}{a_n}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right)\\ \text{ Clearly }{a}_n<0,\text{ if }\frac{1}{a_n}<0\Rightarrow \frac{1}{5}-\frac{4n}{475}+\frac{4}{475}<0\\ \Rightarrow -4n<-99\text{ or }n>\frac{99}{4}=24\frac{3}{4}\therefore n\ge 25\\ \therefore \text{ Least value of }n\text{ is }25.\end{array}$