Which term of the list of the numbers $20, 19\dfrac{1}{4}, 18\dfrac{1}{2}, 17\dfrac{3}{4}, …$ is the first negative term?

The above series is an A.P. with a = 20 and

$d = 19\dfrac{1}{4} - 20 \\[1em] = \dfrac{77}{4} - 20 \\[1em] = \dfrac{77 - 80}{4} \\[1em] = -\dfrac{3}{4}.$

Let nth term be the first negative number.

We know that

a_n = a + (n - 1)d

$\therefore a_n = 20 + (n - 1) \times -\dfrac{3}{4} \\[1em] = 20 - \dfrac{3}{4}n + \dfrac{3}{4} \\[1em] = \dfrac{80 + 3}{4} - \dfrac{3}{4}n \\[1em] = \dfrac{83}{4} - \dfrac{3}{4}n \\[1em]$

Since a_n is a negative number

$\therefore a_n \lt 0 \\[1em] \Rightarrow \dfrac{83}{4} - \dfrac{3}{4}n \lt 0 \\[1em] \Rightarrow \dfrac{3}{4}n \gt \dfrac{83}{4} \\[1em] \Rightarrow n \gt \dfrac{83}{4} \times \dfrac{4}{3} \\[1em] \Rightarrow n \gt \dfrac{83}{3} \\[1em] \Rightarrow n \gt 27\dfrac{2}{3} \\[1em] \therefore n = 28.$

Hence, 28th term is the first negative number in the A.P.