Question

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

Answer 1

$\huge \mathfrak \green{answer}$

Answer 2

Answer:

$\huge\sf\tt\red{AnSwEr}$

Given a1, a2, a3,…are in HP.

So 1/a1, 1/a2, 1/a3 are in AP.

Let d be the common difference,

1/a20 = 1/a1 + 19d

1/a20 – 1/a1 = 19d

1/25- 1/5 = 19d

-4/25 = 19d

d = -4/25×19

a+(n-1)d < 0

=> (1/5) + (n-1)(-4/25×19) < 0

4(n-1)/19×5 > 1

n-1 > 19×5/4

n > 19×5/4 + 1

n = 25

So the least positive value of n is 25.