For the AP 50, 46, 42, 38, ... the terms eventually turn negative, so the running total climbs and then falls.
How many terms give the greatest sum, and what is that greatest sum?

$a=50,d=-4$, so $a_n=54-4n$. The terms stay positive while $54-4n>0$, that is $n<13.5$, so $a_{13}=2$ is the last positive term and $a_{14}=-2$. The sum therefore peaks at $n=13:S_{13}=\frac{13}{2}$$[100+12(-4)]=\frac{13}{2}$(52) = 338. Adding the 14th term drops the total back to 336, and stopping at 12 terms also gives only 336 — both are real totals, just not the maximum.