Let $a_{1}, a_{2}, a_{3}, \ldots . .$ be in harmonic progression with $a_{1}=5$ and $a_{20}=25$. The least…
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
22
23
24
25
Solution
$\because a_{1}, a_{2}, a_{3}, \ldots \ldots$ are in H.P.
$\begin{array}{l}
\therefore \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \ldots \ldots \text { are in A.P. } \\
\therefore \quad \frac{1}{a_{1}}=\frac{1}{5} \text { and } \frac{1}{a_{20}}=\frac{1}{25} \\
\frac{1}{a_{1}}+19 d=\frac{1}{a_{20}} \Rightarrow \frac{1}{5}+19 d=\frac{1}{25} \Rightarrow d=\frac{-4}{475} \\
\text { Now } \frac{1}{a_{n}}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right) \\
\text { Clearly } a_{n} < 0, \text { if } \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5}-\frac{4 n}{475}+\frac{4}{475} < 0 \\
\Rightarrow \quad-4 n < -99 \text { or } n>\frac{99}{4}=24 \frac{3}{4} \quad \therefore n \geq 25 \\
\therefore \quad \text { Least value of } n \text { is } 25 .
\end{array}$