The sum of first five multiples of 3 is

1. 45

2. 55

3. 65

4. 75

First five multiples of 3 are : 3, 6, 9, 12, 15.

The above series is an A.P. with first term = a = 3 and common difference = d = 3.

The formula for sum of A.P. is given by,

$S$n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S5 = \dfrac{5}{2}[2 \times 3 + (5 - 1) \times 3] \\[1em] = \dfrac{5}{2}[6 + 4 \times 3] \\[1em] = \dfrac{5}{2}[6 + 12] \\[1em] = \dfrac{5}{2} \times 18 \\[1em] = 45.

Hence, Option 1 is the correct option.