1) 7, 4, 1, -2. First term is a=7 and common difference is d=-3.
Series is a, a+d, a+2d, ..., a+(n-1)d where n is the term number, starting at n=1.
Sum of the series is Sn=a+(a+d)+(a+2d)+...+(a+(n-1)d).
There are n a's so Sn=an+d(1+2+3+...+(n-1)).
We can pair terms in the series in brackets, which is the sum of the natural numbers up to n-1:
1+2+3+...+n-1=(1+n-1)+(2+n-2)+(3+n-3)+...=n+n+n+...
There are n-1 terms so there are (n-1)/2 pairs and each pair sums to n so the sum of the natural numbers up to n-1 is n(n-1)/2, because each pair sums to n. By the way, it doesn't matter whether n is odd or even, the formula still works.
So the sum of the given series is an+dn(n-1)/2=n(a+d(n-1)/2). Now put a=7, n=20 and d=-3:
S20=140-60×19/2=140-570=-430.
Let's check if the formula is correct by summing the given 4 terms: 7+4+1-2=10. Now put n=4:
S4=28-12(3)/2=28-18=10. So the formula is correct.
20th term=7+19(-3)=7-57=-50.
2) Since the nth term is a+(n-1)d
a=5, an=17, Sn=44.
a+(n-1)d=17=5+(n-1)d ①; an+dn(n-1)/2=44=5n+dn(n-1)/2=n(5+d(n-1)/2) ②.
We have two equations and two unknowns so we can find the unknowns.
We know from ① that 5+(n-1)d=17, so (n-1)d=12 and (n-1)d/2=6, which can be substituted in ②
n(5+6)=44, 11n=44, n=4. We can now use ① to find d: 5+3d=17, 3d=12, d=4.
The series is 5, 9, 13, 17 (an=a4=17); Sn=S4=5+9+13+17=44.
3) Geometric series 4, 12, 36, 108, ... a=4, common ratio r=3.
s10=a(rn-1)/(r-1)=4(310-1)/2=2(59049-1)=118096.