jmc

The arithmetic series is $1+4+\cdots +25+28$, with a common difference of 3. Suppose there are $n$ terms in the series. Then 28 is the $n$th term, so $1+(n-1)\cdot 3=28$. Solving, we get $n=10$. The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so the sum is $(1+28)/2\cdot 10=\boxed{145}$.