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AIEEE 2012

JEE Main 2012 Mathematics question on Sequences and Series

If a1,a2,a3,…a_1, a_2, a_3, \ldots are in a harmonic progression with a1=5a_1 = 5 and a20=25a_{20} = 25. Then, the least positive integer nn for which an<0a_n < 0, is
  1. A.
    22
  2. B.
    23
  3. C.
    24
  4. D.
    25
Show answer and solution

Correct option: D

Solution

In HP, the reciprocals 1/a₁, 1/a₂, ... form an AP. Here 1/a₁ = 1/5 and 1/a₂₀ = 1/25.

Common difference d = (1/25 - 1/5)/19 = (1/25 - 5/25)/19 = (-4/25)/19 = -4/475.

The nth term: 1/aₙ = 1/5 + (n-1)(-4/475) = 1/5 - 4(n-1)/475.

For aₙ < 0, we need 1/aₙ < 0:
1/5 - 4(n-1)/475 < 0
95/475 < 4(n-1)/475
95 < 4(n-1)
n-1 > 95/4 = 23.75
n > 24.75

The least positive integer n is 25.

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