Let $a_1 , a_2, a_3$ be a harmonic progression with $a_1 =5$ and $a_{20} =25$. Then the least positive integer for which $ a_n \lt 0$ is?

Question

Let $a_1 , a_2, a_3$ be a harmonic progression with $a_1 =5$ and $a_{20} =25$. Then the least positive integer for which $ a_n \lt 0$ is?

My attempt:- As This is a harmonic progression,

so the denominator of the first term is

$a=\frac{1}{5}$

and the denominator
of $a_{20}=a+19d=\frac{1}{25}$
solving the above linear equations we find that $d=\frac{-4}{475}$

so finding when the denominator will be zero we get $n=23.75$, as the closest integer is 24, the HP becomes negative when $n=24$

However, my book says that the answer is actually $25$ and I'm not sure I understand how.

Where Have I gone wrong?

Thanks in advance for the help!

Source:- JEE Advance , Paper 2,2012

Answer 1

What you get is $n-1=24$. Recall that $a_n=\frac{1}{a+(n-1)d}$.