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Let `a_(1),a_(2),a_(3), . . .` be a harmonic progression with `a_(1)=5anda_(20)=25`. The least positive integer n for which `a_(n)lt0`, is

A

22

B

23

C

24

D

25

Text Solution

Verified by Experts

The correct Answer is:
d
Here, `a_(1) =5, a_(20) = 25` for HP.
` :. 1/a=5 and 1/(a+19d) =25`
` rArr 1/5 + 19d = 1/25`
`rArr d = (-4)/(19 xx25)`
Since, ` a_(n) lt 0 rArr 1/5 +(n-1)d lt 0`
` 1/5 - 4/(19 xx25) (n-1) lt 0 `
` rArr (n-1) gt 95/4`
` rArr n gt 1 +95/4`
or ` n gt 24.75`
Least positive value of n = 25.
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