Some integers between 100 and 300 leave remainder 2 when divided by 9. What is the sum of all such integers?

Such integers are 101, 110, 119, ..., 299 — an AP with $a=101,d=9$ and $l=299$. Then $n=\frac{299-101}{9}+1=22+1=23$, so $S=\frac{23}{2}$$(101+299)=23(200)=4600$. The value 4400 uses $n=22$, the classic slip of forgetting the "+1" when counting terms. 4800 uses $n=24$. 9200 omits the factor $\frac{1}{2}$.