Question 1103960
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The first term is 3 and the common difference is 4; so the 80th term is 3+79*4 = 319.<br>
Let's write the beginning and end of the series of numbers:
3+7+11+15+ ... +311+315+319<br>
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), ....<br>
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.