Find the sum

6+9+12 + ... + 93

Sn=

We notice that:

9-6 = 3

12-9 = 3, Then we can conclude that this is an arithmetic progression with:

1st term a₁ = 6 and with common difference d = 3

So the first term is 6 and the last term is 93.

the formula of the sum in a A. P is:

S = (a + last term). n/2, n being the rank of the last term. So to be

able to find S, we have to calculate the value of n

We know that the last value of a A. P is:

last value = a₁ + (n-1) d

93 = 6 + (n-1) (3) → 93 = 6 + 3n - 3 → n = 90/3 → n = 30 (rank 30th)

Now we can find the sum:

S = (a₁ + last term) n/2

S = (6+93) 30/2

S = (99).15 = 1,485