9-3 Geometric Sequences & Series

Geometric Sequence The ratio of a term to it’s previous term is constant. This means you multiply by the same number to get each term. This number that you multiply by is called the common ratio (r).

Example: Decide whether each sequence is geometric. 4,-8,16,-32,… -8/4=-2 16/-8=-2 -32/16=-2 Geometric (common ratio is -2) 3,9,-27,-81,243,… 9/3=3 -27/9=-3 -81/-27=3 243/-81=-3 Not geometric

Find the rule for an for the following sequence. 2, 4, 8, 16, 32… 1st, 2nd, 3rd, 4th, 5th Think of how to use the common ratio, n and a1, to determine the term value.

Rule for a Geometric Sequence an=a1rn-1 Example 1: Write a rule for the nth term of the sequence 5, 20, 80, 320,… . Then find a8. First, find r. r= 20/5 = 4 an=5(4)n-1 a8=5(4)8-1 a8=5(4)7 a8=5(16,384) A8=81,920

Example 2: One term of a geometric sequence is a4=3 Example 2: One term of a geometric sequence is a4=3. The common ratio is r=3. Write a rule for the nth term. Use an=a1rn-1 3=a1(3)4-1 3=a1(3)3 3=a1(27) 1/9=a1 an=a1rn-1 an=(1/9)(3)n-1

Ex 3: Two terms of a geometric sequence are a2=-4 and a6=-1024 Ex 3: Two terms of a geometric sequence are a2=-4 and a6=-1024. Write a rule for the nth term. Write 2 equations, one for each given term. a2=a1r2-1 OR -4=a1r a6=a1r6-1 OR -1024=a1r5 Use these 2 equations & substitution to solve for a1 & r. -4/r=a1 -1024=(-4/r)r5 -1024=-4r4 256=r4 4=r & -4=r If r=4, then a1=-1. an=(-1)(4)n-1 If r=-4, then a1=1. an=(1)(-4)n-1 an=(-4)n-1 Both Work!

Formula for the Sum of a Finite Geometric Series n = # of terms a1 = 1st term r = common ratio

Example 4: Consider the geometric series 4+2+1+½+… . Find the sum of the first 10 terms. Find n such that Sn=31/4.

log232=n

Looking at infinite series, what happens to the sum as n approaches infinity in each case? Notice, if and thus the sum does not exist.

So what if Looking at infinite series, what happens to the sum as n approaches infinity if ?

Sum of a Infinite Geometric Series when n = # of terms a1 = 1st term r = common ratio

Ex 5: Find the Sum of the infinite series a) 1 + 1.5 + 2.25 + 3.375 + … Sum DNE since r = 1.5 and is > 1 b) 9 + 6 + 4 + 8/3 + … r = 2/3 and is < 1 so we use the formula

H Dub 9-3 Pg.669 #3-42 (3n), 53-55, 73-75, 79-81