In the sum of the series

$S_{n}=20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$

$\because S_{n}=488$

$488=\frac{n}{2}\left[2\left(\frac{100}{5}\right)+(n-1)\left(-\frac{2}{5}\right)\right]$

$488=\frac{n}{2}(101-n) \Rightarrow n^{2}-101 n+2440=0$

$\Rightarrow n=61$ or 40

For $n=40 \Rightarrow T_{n}>0$

For $n=61 \Rightarrow T_{n}<0$

$n^{\text {th }}$ term $=T_{61}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$