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 [2012]

Question

Let $a_{1}, a_{2}, a_{3}, \dots$ be in harmonic progression with $a_{1} = 5$ and $a_{20} = 25$ . The least positive integer $n$ for which $a_{n} < 0$ is [2012]

Smart Achievers · Accepted Answer

(4)

$∵$ $a_{1}, a_{2}, a_{3}, \dots are in H.P.$

$∴$ $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \dots are in A.P.$

$∴$ $\frac{1}{a_{1}} = \frac{1}{5} and \frac{1}{a_{20}} = \frac{1}{25}$

$\frac{1}{a_{1}} + 19 d = \frac{1}{a_{20}} \Rightarrow \frac{1}{5} + 19 d = \frac{1}{25} \Rightarrow d = \frac{- 4}{475}$

$Now \frac{1}{a_{n}} = \frac{1}{5} + (n - 1) (\frac{- 4}{475})$

$Clearly a_{n} < 0, if \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5} - \frac{4 n}{475} + \frac{4}{475} < 0$

$\Rightarrow - 4 n < - 99$ or $n > \frac{99}{4} = 24 \frac{3}{4}$ $∴$ $n \geq 25$

$∴$ $Least value of n is 25 .$

Solution

Q.
Let $a_{1}, a_{2}, a_{3}, \dots$ be in harmonic progression with $a_{1} = 5$ and $a_{20} = 25$ . The least positive integer $n$ for which $a_{n} < 0$ is [2012]

1 22

2 23

3 24

4 25

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