The first term is 3 and the common difference is 4; so the 80th term is 3+79*4 = 319.
Let's write the beginning and end of the series of numbers:
3+7+11+15+ ... +311+315+319
Because the terms in the sequence are equally spaced, the 80 terms in the sequence can be grouped into 40 pairs, each with the same sum:
(3+319), (7+315), (11+311), ....
So the sum of all 80 terms is the sum of 40 pairs, each with a sum of (3+319)=322. That means the sum of the 80 terms is 322*40 = 12880.
Arithmetic progression with the first term = 3 and the common difference d = 4.
Use the general formula for the sum of the first n terms of an arithmetic progression
= = .
In your case it is = = 12880.
