If the sum of the series$20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+\dots \text{ up to }{n}^{\text{th }}\text{ term is }488\text{ and the }{\mathrm{n}}^{\text{th }}$ term is negative, then :

$\text{ The given sum is }S=20+\left(19+\frac{3}{5}\right)+\left(19+\frac{1}{5}\right)+\dots .n\text{ terms }$

$\text{ The terms are in arithmetic progression with first term }a=20,\text{ and the common difference }d=-\frac{2}{5}$

$\text{ Hence, sum of first }n\text{ terms is }S=\frac{n}{2}(2a+(n-1\left)d\right)$

Given 

$\begin{array}{c}488=\frac{n}{2}\left[2\left(20\right)+\left(n-1\right)\left(\frac{-2}{5}\right)\right]\\ =n\left(20+\frac{1-n}{5}\right)\\ =\frac{1}{5}\left(100n+n-n^2\right)\end{array}$

It gives 

$n^2-101n+5\times 488=0$

$\text{ Hence, }n=40\text{ (or) }61$

$\text{ Since the }n\text{ th term is negative, }$

$\begin{array}{c}{t}_n=a+\left(n-1\right)d\\ =20+\left(61-1\right)\left(-\frac{2}{5}\right)\\ =20+60\left(-\frac{2}{5}\right)\\ =20-24\\ =-4\end{array}$

$\text{ Therefore, }n=61,t_n=-4$