If you know the formula of a given arithmetic series, it is (first term + last term)*(number of terms)/2. we already know we want the first 30 terms so number of terms = 30. now we should try to find the first and last term. we can do that by finding the d, or the distance between each successive term. this is: (116 - 74)/(12 - 5) = 6. now we can find the first term. the fourth term is 74 - 6, the third term is 74 - 2(6), the second term is 74 - 3(6) and the first term is 74 - 4(6) = 50. now using the first term, we can find the last term. the last term is the 30th term and to get that number, some might think it is: 50 + 30(6) but take a look at the second term, it is 50 + 6 and the third term, it is 50 + 2(6) basically the nth term is 50 + (n-1)(6) so the 30th term is: 50 + 29(6) = 224 now we can compute what we want: (224 + 50)(30)/2 = 4110
now you may be wondering why the formula is the way it is. there is actually a story associated with this. there was a famous mathematician named carl friedrich gauss. legend has it that when he was a kid, once when he was in school, the teacher wanted to slack off, so he (or she...idk which one) told all his (her) students to add the numbers from 1 to 100. everyone instantly went 1+2 = 3, 3 + 3 = 6, 6 + 4 = 10, etc. carl, on the other hand, applied a neat trick that got him the answer really quickly. he wrote the integers from 1 to 100 then wrote that same sequence below it but in reverse order. so: 1, 2, 3, 4, 5, ..., 95, 96, 97, 98, 100 100, 99, 98, 97, 96, 95...5, 4 3, 2, 1 now add the first term of the top sequence to the first term of the bottom sequence (1 + 100 = 101), and do the same thing for the 2nd term of top and bottom sequence, and do it for all 100 terms. you'll find that each term adds to 101. so the total sum of the 2 identical sequences is (101)(100). the 101 is the first term+last term while the 100 is the number of terms (starting to look like the formula i proposed in the beginning!). now remember this is the sum of 2 sequences but we only want the sum of 1. since the sum of the 2 sequences are the same (because they have the same number but written in different orders) we can just take (101)(100) and divide that by 2 to get the sum of the first 100 positive integers. this can be extended to any arithmetic sequence like so: a, a+d, a+2d... a+(n-2)d, a+(n-1)d a+(n-1)d, a+(n-2)d, ...a+d, a a is the first term of the sequence and d is the distance and the sequence has n terms. each of the n pairs sum to a + (a+(n-1)d) and we can do the same thing as carl did to get the formula i used to solve this problem.
