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(D) In a harmonic progression $(HP)$,the reciprocals of the terms form an arithmetic progression $(AP)$.Let the $AP$ be $b_1, b_2, b_3, \dots$ where $

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$25$

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(D) In a harmonic progression $(HP)$,the reciprocals of the terms form an arithmetic progression $(AP)$.Let the $AP$ be $b_1, b_2, b_3, \dots$ where $b_n = \frac{1}{a_n}$.Given $a_1 = 5 \implies b_1 = \frac{1}{5}$ and $a_{20} = 25 \implies b_{20} = \frac{1}{25}$.The formula for the $n$-th term of an $AP$ is $b_n = b_1 + (n - 1)d$.For $n = 20$: $\frac{1}{25} = \frac{1}{5} + (20 - 1)d$.$\frac{1}{25} - \frac{1}{5} = 19d \implies \frac{1 - 5}{25} = 19d \implies -\frac{4}{25} = 19d \implies d = -\frac{4}{475}$.We want $a_n $b_n = \frac{1}{5} + (n - 1)(-\frac{4}{475}) $\frac{1}{5} Multiply by $475$: $95 $23.75 $n > 24.75$.The smallest positive integer $n$ is $25$.

Let $a_1, a_2, a_3, \dots$ be a harmonic progression where $a_1 = 5$ and $a_{20} = 25$. What is the smallest positive integer $n$ such that $a_n < 0$?

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