Let a_1, a_2, a_3, ldots be in harmonic progr | vedclass

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(D) Since $a_1, a_2, \ldots$ are in harmonic progression,their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \ldots$ are in arithmetic progression.Let th

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$25$

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(D) Since $a_1, a_2, \ldots$ are in harmonic progression,their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \ldots$ are in arithmetic progression.Let the arithmetic progression be $b_n = \frac{1}{a_n} = A + (n-1)D$.Given $a_1 = 5 \Rightarrow b_1 = \frac{1}{5}$ and $a_{20} = 25 \Rightarrow b_{20} = \frac{1}{25}$.$b_{20} = b_1 + 19D \Rightarrow \frac{1}{25} = \frac{1}{5} + 19D$.$19D = \frac{1}{25} - \frac{1}{5} = \frac{1-5}{25} = -\frac{4}{25}$.$D = -\frac{4}{19 	imes 25} = -\frac{4}{475}$.We want $a_n  0$ and $a_{20} > 0$,we look for when the term becomes negative.$b_n = \frac{1}{5} - (n-1)\frac{4}{475} $\frac{1}{5} $\frac{475}{20} $23.75 $n > 24.75$.The least positive integer $n$ is $25$.

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

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