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

This would be an arithmetic course.

Lets original observe at the original three stipulations of this course.

3, 6, 9

What do you attend-to?

Each compute succeeding the original is 3 larger than the earlier compute.

This signifies the progression is an arithmetic course.

To meet the sum of an arithmetic course, there are two unconnected formulas. The one we earn use is ##s_n = n/2(2a + (n - 1)d)##

n, or the compute of stipulations, is 30. d, the beggarly dissonance, is 3. a, the original expression in the course is 3.

Plugging these computes into the formula, we get:

##s_30 = 30/2(2(3) + (29)3)##

##s_30 = 15(6 + 87)##

##s_30 = 15(93)##

##s_30 = 1395##

**Practice exercises:**

- Find the sum of the original 12 dogmatical multiples of 13.

2 . Meet the sum of all the orderly plain integers in betwixt the 4th meanest dogmatical multiple of 16 and the 9th meanest dogmatical multiple of 22.**Challenge Problem**

In an arithmetic course, the original compute is ##n##, the promote ##n + 5##, the third ##n+10##. Meet the sum of the original 38 stipulations in this course.

Good luck!

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