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

(d) : We have

\frac{1}{a_{20}}=\frac{1}{a_1}+19 d \Rightarrow \frac{1}{25}=\frac{1}{5}+19 d

\n\Rightarrow d=\frac{1}{19}\left(\frac{1}{25}-\frac{1}{5}\right)=-\frac{4}{19 \times 25}\n

\n\nAgain

a_n<0

\n\begin{aligned}\n& \Rightarrow \frac{1}{5}+(n-1)\left(-\frac{4}{19 \times 25}\right)<0 \\\n& \frac{(n-1) 4}{19 \times 5}>1 \Rightarrow n>\frac{19 \times 5}{4}+1 \\\n& \Rightarrow n \geq 25\n\end{aligned}\n

\n\nThus 25 is the least value of

n

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