jmc

The first ten positive multiples of 3 are 3, $3\cdot 2$, $\dots$, $3\cdot 10$, so we want to find the sum $3+3\cdot 2+\cdots +3\cdot 10=3\cdot (1+2+\cdots +10)$.

For all $n$, $1+2+\cdots +n=n(n+1)/2$, so $3\cdot (1+2+\cdots +10)=3\cdot 10\cdot 11/2=\boxed{165}$.