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

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Solution

The correct option is D 25
$∵ a_{1}, a_{2}, a_{3}, . . . .$ are in H.P.
$∴ \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} . . . . . . .$ are in A.P.
$∴ \frac{1}{a_{1}} = \frac{1}{5} a n d \frac{1}{a_{20}} = \frac{1}{25}$
$\frac{1}{a_{1}} + 19 d = \frac{1}{a_{20}} \Rightarrow \frac{1}{5} + 19 d = \frac{1}{25} \Rightarrow d = \frac{- 4}{475}$
Now $\frac{1}{a_{n}} = \frac{1}{5} + (n - 1) (\frac{- 4}{475})$
Clearly $a_{n} < 0 i f \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5} - \frac{4 n}{475} + \frac{4}{475} < 0$
$\Rightarrow - 4 n < - 99 o r n > \frac{99}{4} = 24 \frac{3}{4} ∴ n \geq 25$
Hence least value of n is 25.

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Let a1,a2,a3...... be in harmonic progression with a1=5 and a20=25. The least positive integer n for which an<0 is

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