# How do you find the sum of the first 30 positive multiples of 3?

Mar 4, 2016

This would be an arithmetic series.

#### Explanation:

Lets first look at the first three terms of this series.

3, 6, 9

What do you notice?

Each number after the first is 3 larger than the previous number.

This signifies the sequence is an arithmetic series.

To find the sum of an arithmetic series, there are two separate formulas. The one we will use is ${s}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

n, or the number of terms, is 30. d, the common difference, is 3. a, the first term in the series is 3.

Plugging these numbers into the formula, we get:

${s}_{30} = \frac{30}{2} \left(2 \left(3\right) + \left(29\right) 3\right)$

${s}_{30} = 15 \left(6 + 87\right)$

${s}_{30} = 15 \left(93\right)$

${s}_{30} = 1395$

Practice exercises:

1. Find the sum of the first 12 positive multiples of 13.

2 . Find the sum of all the consecutive even integers in between the 4th smallest positive multiple of 16 and the 9th smallest positive multiple of 22.
Challenge Problem

In an arithmetic series, the first number is $n$, the second $n + 5$, the third $n + 10$. Find the sum of the first 38 terms in this series.

Good luck!