The sum of the first first five multiples of 3 is

The first five multiples of 3 are 3, 6, 9, 12, and 15.
Summing these numbers directly:

$$\text{Sum}=3+6+9+12+15=45$$

Alternatively, using the sum formula of an Arithmetic Progression ($\text{AP}$) where $n=5$, first term $a=3$, and last term $l=15$:

$$\text{Sum}=\frac{n}{2}(a+l)=\frac{5}{2}(3+15)=\frac{5}{2}(18)=5\times 9=45$$