Which term of the AP 20,

The given $\mathrm{AP}$ is $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .$

Here, $a=20$ and $d=19 \frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$

Let the nth term of the given AP be the first negative term. Then,

$a_{n}<0$

$\Rightarrow 20+(n-1) \times\left(-\frac{3}{4}\right)<0 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 20+\frac{3}{4}-\frac{3}{4} n<0$

$\Rightarrow \frac{83}{4}-\frac{3}{4} n<0$

$\Rightarrow-\frac{3}{4} n<-\frac{83}{4}$

$\Rightarrow n>\frac{83}{3}=27 \frac{2}{3}$

$\therefore n=28$

Hence, the 28th term is the first negative term of the given AP.