The sum of first five multiples of 3 is:

We have to find the sum of first five multiples of 3

First five multiples of 3 are 3, 6, 9, 12 and 15

It forms an Arithmetic Progression (A.P) with

First term 

Common difference, $d=a_{2}-a$

$=6-3$

$=3$

We know that the sum of the $n$ terms of an arithmetic progression $\mathrm{S}_{n}=\frac{n}{2}(2 a+(n-1) d)$

Here $n=5$

Therefore

$\mathrm{S}_{5}=\frac{5}{2}(2 \times 3+(5-1) 3)$

$=\frac{5}{2} \times 18$

$=45$

Hence option (a) is correct.