Given sequence is $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$
Which can be rewritten as $20, \frac{77}{4}, \frac{37}{2}, \frac{71}{4}, \ldots$
This is an AP series
Here, first term $\mathrm{a}=20$ and common difference $\mathrm{d}=-\frac{3}{4}$
$\mathrm{n}^{\text {th }} \operatorname{term}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=20+(n-1)\left(-\frac{3}{4}\right)$
$=\frac{83}{4}-\frac{3}{4} n$
For first negative term, $n^{\text {th }}$ term $ < 0$
$\Rightarrow \frac{83}{4}-\frac{3}{4} n < 0 \Rightarrow 83 < 3 n$
$\Rightarrow n>\frac{83}{3}=27.66$
So, $\mathrm{n}$ should be 28 .
Hence, 28 th term is first negative term.