The given series is an arithmetic progression whose first term is 20 and common difference is 1953−20=−52.
According to question,
488=2n[2(20)+(n−1)(−52)]
⇒488=2n(40−52n+52)⇒n2−101n+2440=0
⇒n=61 or 40
For n=40⇒Tn=T40=20+(40−1)(−52)=522>0
For n=61⇒Tn=T61=20+(61−1)(−52)=−4<0
Thus, Tn=T61=−4