1. Find S20 For Each Arithmetic Series Listed Below. A) 70 + 66 + 62 + 58 ... = 256 B) 3 + 9 + 15 + 21... = 48 C) 9 + 16 + 23 ... = 48 D) 3 + 2.5 + 2 + 1.5 ... = 9?

A. The Nth term is given by 74 - 4N, so the sum of 20 terms is
  20*74 - 4*(20)(21)/2 = 1480 - 840 = 640

B. The Nth term is given by -3 + 6N, so the sum of 20 terms is
  20*(-3) + 6(20)(21)/2 = -60 + 1260 = 1200

C. The Nth term is given by 2 + 7N, so the sum of 20 terms is
  20*2 + 7*210 = 40 + 1470 = 1510

D. The Nth term is given by 3.5 - 0.5N, so the sum of 20 terms is
  20*3.5 - 0.5(210) = 70 - 105 = -35

---

The sum of the numbers 1 to n is (n)(n+1)/2. This means the sum of the numbers 1-20 is
  (20)(20+1)/2 = 10*21 = 210
The sum of the numbers k*1 to k*n will be k(n)(n+1)/2, where k is some constant multiplier.

The sum of an arithmetic series of 20 terms is 10 times the sum of the first term and the 20th term. (The 2nd and 19th terms add to the same sum, and so on.) The series in part A, for example, has a first term of 70 and a 20th term of -6. The sum of those is 64, so the sum of the first 20 terms is 64*10 = 640.