AMC 12 Mock Exam 3 Problem 11

Let the first term be $a$ and the common difference $d$. Then $5(2a+9d)=155$ and $10(2a+19d)=610$, i.e., $2a+9d=31$ and $2a+19d=61$, so $d=3$ and $a=2$; the $n$th term is $3n-1$.

The terms $21$ through $30$ sum to $\sum_{n=21}^{30}(3n-1)=3(21+22+\cdots+30)-10=3\cdot255-10=\boxed{755}$.

(Alternatively, the sums of consecutive blocks of $10$ terms form an arithmetic sequence: $155$, $610-155=455$, so the next block is $755$.)