If the sum of the series 20 + 19 3 5 + 19 1 5 + 18 4 5 + . . . . . . . . . . up to n th  term is 488…

The given series is an arithmetic progression whose first term is $20$ and common difference is $19\frac{3}{5}-20=-\frac{2}{5}$.

According to question,

$488=\frac{n}{2}\left[2\left(20\right)+\left(n-1\right)\left(-\frac{2}{5}\right)\right]$

$\Rightarrow 488=\frac{n}{2}\left(40-\frac{2n}{5}+\frac{2}{5}\right)\Rightarrow n^2-101n+2440=0$

$\Rightarrow n=61$ or $40$

For $n=40\Rightarrow T_n=T_{40}=20+\left(40-1\right)\left(-\frac{2}{5}\right)=\frac{22}{5}>0$

For $n=61\Rightarrow T_n=T_{61}=20+\left(61-1\right)\left(-\frac{2}{5}\right)=-4<0$

Thus, $T_n=T_{61}=-4$