The sum of first five multiples

(a) The first five multiples of 3 are 3, 6, 9,12 and 15.

Here, first term, a = 3, common difference, d = 6-3 = 3 and number of terms, n = 5

$\therefore \quad S_{5}=\frac{5}{2}[2 a+(5-1) d] \quad\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$

$=\frac{5}{2}[2 \times 3+4 \times 3]$

$=\frac{5}{2}(6+12)=5 \times 9=45$