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Q.
If the sum of the series $20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots $ upto $n^ {th }$ term is $488$ and the $n ^ {th }$ term is negative, then :
$S =\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\frac{94}{5}+\ldots n$
$S _{ n }=\frac{ n }{2}\left(2 \times \frac{100}{5}+( n -1)\left(-\frac{2}{5}\right)\right)=188$
$n (100- n +1)=488 \times 5$
$n ^{2}-101 n +488 \times 5=0$
$n =61,40$
$T _{ n }= a +( n -1) d =\frac{100}{5}-\frac{2}{5} \times 60$
$=20-24=-4$