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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
`a_(1),a_(2),a_(3),…..` are in H.P.
`rArr1/a_(1),1/a_(2),1/a_(3),`… are in A.P.
`rArr1/a_(n)=1/a_(1)+(n-1)dlt0`
where `(1/25-5/25)/19=d=((-4)/(19xx25)) `
`rArr1/5+(n-1)((-4)/(19xx25))lt0`
or `(4(n-1))/(19xx5)gt1`
or `n-1gt(19xx5)/4`
or `ngt(19xx5)/4`+1
or `nge25`
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