Precal question?

Question

find the sum of each geometric sequence.
a.{36,12,4,...4/27}
b.{20,10,5,...5/16}

Answer 1

The sum of a geometric sequence ∑(i=0, ∞, A*s^i) is A / (1 - s) where s is the ratio of one term to its previous one and |s| < 1. A finite series of n terms can be thought of as the superposition of two such infinite series, namely,
∑(i=0, ∞, A*s^i) - ∑(i=n, ∞, A*s^i)
= ∑(i=0, ∞, A*s^i) - s^n ∑(i=0, ∞, A*s^i)
= A (1 - s^n) / (1 - s)
= (A - As^n) / (1 - s)
where A represents the first term of the series and As^n represents the first term beyond the last term of the finite series.

For series a, A = 36, s = 1/3, and As^n = 4/81. Therefore, the sum of this series would be
(36 - 4/81) / (1 - 1/3)
= 1456 / 27

I will leave series b to you, though I will give you the hint that
A = 20, s = 1/2, and As^n = 5/32.

Answer 2

Question a
36 , 12 , 4 , 4 / 3 , 4 / 9 , 4 / 27
6 terms
r = 1/3
a = 36
S6 = a(1 - r^6) / (1 - r)
S6 = (36)(1 - 1/729) / (2/3)
S6 = (36)(728/729) (3/2)
S6 = (4)(728/81) (3/2)
S6 = (6)(728/81)
S6 = 2 (728/27)
Question b
a = 20
r = 1/2
6 terms
S6 = (20)(1 - 1/64) / (1/2)
S6 = (20)(63/64) / (1/2)
S6 = 40(63/64)
S6 = (5/8)(63)
S6 = 315 / 8