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

a1,a2,a3..are in H.P. ⇒1/a1, 1/a2, 1/a3…are in A.P. ⇒1/an = 1/a1 + (n-1)d ⇒1/a20 = 1/a1 + 19d ⇒d = 1/25 - 1/5 = -4/25
Given that an < 0 ⇒1/an < 0 ⇒1/a1 + (n-1)(-4/25) < 0 ⇒1/5 + (n-1)(-4/25) < 0 ⇒(n-1)(-4/25) < -1/5 ⇒(n-1)(4/25) > 1/5 ⇒(n-1) > 25/4 ⇒n > 25/4 + 1 ⇒n ≥ 7
1/an = 1/5 + (n-1)(-4/25) < 0
1/5 - 4(n-1)/25 < 0
5 - 4(n-1) < 0
5 - 4n + 4 < 0
9 - 4n < 0
4n > 9
n > 9/4 = 2.25
Since n must be an integer, the least positive integer value of n is 3.
However, this contradicts the given solution. Let's re-examine the calculation.
1/an = 1/a1 + (n-1)d
1/a20 = 1/a1 + 19d
1/25 = 1/5 + 19d
19d = 1/25 - 1/5 = -4/25
d = -4/475
1/an = 1/5 + (n-1)(-4/475) < 0
1/5 < (n-1)(4/475)
475/20 < n-1
23.75 < n-1
n > 24.75
Therefore, the least positive integer n is 25.