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

Let a₁, a₂, a₃, … be in harmonic progression with a₁ = 5 and a₂₀ = 25. The least positive integer n for which aₙ < 0 is $\text{Given: } a_1, a_2, a_3, \ldots \text{ is an H.P.}$ $a_1 = 5$ $a_{20} = 25$ $\text{To find: least } n \text{ such that } a_n < 0$ $\text{H.P.} \iff \dfrac{1}{a_1}, \dfrac{1}{a_2}, \dfrac{1}{a_3}, \ldots \text{ is an A.P.}$ $b_n = \dfrac{1}{a_n} \quad\Rightarrow\quad b_n \text{ is an A.P.}$ $b_1 = \dfrac{1}{a_1} = \dfrac{1}{5}$ $b_{20} = \dfrac{1}{a_{20}} = \dfrac{1}{25}$ $d = \dfrac{b_{20} - b_1}{20 - 1}$ $d = \dfrac{\frac{1}{25} - \frac{1}{5}}{19} = \dfrac{-\frac{4}{25}}{19}$ $d = -\dfrac{4}{475}$ $b_n = b_1 + (n-1)d$ $b_n = \dfrac{1}{5} + (n-1)\left(-\dfrac{4}{475}\right)$ $b_n = \dfrac{95 - 4(n-1)}{475} = \dfrac{99 - 4n}{475}$ $a_n < 0 \iff \dfrac{1}{a_n} < 0 \iff b_n < 0$ $\dfrac{99 - 4n}{475} < 0 \Rightarrow 99 - 4n < 0$ $4n > 99$ $n > \dfrac{99}{4} = 24.75$ $\text{least positive integer } n = 25$ $a_{24} = \dfrac{475}{99 - 96} = \dfrac{475}{3} > 0$ $a_{25} = \dfrac{475}{99 - 100} = -475 < 0$ $\boxed{n = 25} \quad \text{option (4)}$