Mathematics - Sequence & Series Question with Solution | TestHub

It is given that $a_1,a_2,a_3\dots$. are in HP. Therefore, $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\dots \dots .$. are in AP. Let $d$ be the common difference of the AP. $\therefore \frac{1}{a_n}=\frac{1}{a_1}+(n-1)d$ and $\frac{1}{a_{20}}=\frac{1}{a_1}+19d$ $\Rightarrow \frac{1}{a_n}=\frac{1}{5}+(n-1)d$ and $\frac{1}{25}=\frac{1}{5}+19d$ $\Rightarrow \frac{1}{a_n}=\frac{1}{5}+(n-1)d$ and $d=\frac{-4}{19\times 25}$ $\Rightarrow \frac{1}{a_n}=\frac{1}{5}-\frac{4(n-1)}{19\times 25}$ $\Rightarrow \frac{1}{a_n}=\frac{95-4n+4}{19\times 25}$ $\Rightarrow a_n=\frac{99-4n}{19\times 25}$ Now, $a_n<0$ $\Rightarrow \frac{99-4n}{19\times 25}<0$ $\Rightarrow 99-4n<0$ $\Rightarrow 4n>99$ $\Rightarrow n>24\frac{3}{4}\Rightarrow n\ge 25$