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9-3 Geometric Sequences 9-3 Geometric Sequences & Series& Series
Geometric SequenceGeometric Sequence
• The ratio of a term to it’s previous term The ratio of a term to it’s previous term is constant.is constant.
• This means you multiply by the same This means you multiply by the same number to get each term.number to get each term.
• This number that you multiply by is This number that you multiply by is called the called the common ratiocommon ratio (r). (r).
ExampleExample: Decide whether each : Decide whether each sequence is geometric.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 Find the rule for aan n for the for the
following sequencefollowing sequence..• 2, 4, 8, 16, 32… 1st, 2nd, 3rd, 4th, 5th
Think of how to use the common ratio, n
and aa1, 1, to determine
the term value.
Rule for a Geometric SequenceRule for a Geometric Sequence
aann=a=a11rrn-1n-1
Example 1Example 1: Write a rule for the nth term of the : Write a rule for the nth term of the sequence 5, 20, 80, 320,… . Then find asequence 5, 20, 80, 320,… . Then find a88..
•First, find r.First, find r.
•r= r= 2020//5 5 = 4= 4
•aann=5(4)=5(4)n-1n-1
aa88=5(4)=5(4)8-18-1
aa88=5(4)=5(4)77
aa88=5(16,384)=5(16,384)
AA88=81,920=81,920
Example 2Example 2: One term of a geometric sequence : One term of a geometric sequence is ais a44=3. The common ratio is r=3. Write a rule =3. The common ratio is r=3. Write a rule
for the nth term.for the nth term.
• Use aUse ann=a=a11rrn-1n-1
3=a3=a11(3)(3)4-14-1
3=a3=a11(3)(3)33
3=a3=a11(27)(27)11//99=a=a11
• aann=a=a11rrn-1n-1
aann=(=(11//99)(3))(3)n-1n-1
Ex 3: Two terms of a geometric sequence are aEx 3: Two terms of a geometric sequence are a22=-4 =-4
and aand a66=-1024. Write a rule for the nth term.=-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 Both Work!Work!
Formula for the Sum of a Finite Formula for the Sum of a Finite Geometric SeriesGeometric Series
r
raS
n
n 1
11
n = # of termsn = # of terms
aa1 1 = 1= 1stst term term
r = common ratior = common ratio
Example 4: Consider the geometric Example 4: Consider the geometric series 4+2+1+½+… .series 4+2+1+½+… .
• Find the sum of the first 10 terms.
• Find n such that Sn=31/4.
r
raS
n
n 1
11
21
1
21
14
10
10S
128
1023
1024
20464
21
10241023
4
2110241
1410
S
21
1
21
14
4
31
n
21
1
21
14
4
31
n
2121
14
4
31
n
n
2
118
4
31
n
2
11
32
31n
2
1
32
1n
2
1
32
1
5n
n
n
2
1
32
1
n232 log232=n
Looking at infinite series, what Looking at infinite series, what happens to the sum as n happens to the sum as n
approaches infinity in each approaches infinity in each case? case?
3 + 9 + 27 + 81, +….+ 3n3 + 9 + 27 + 81, +….+ 3n
27 + 9 + 3, + 1 + 1/3 + ….+ (1/3)n27 + 9 + 3, + 1 + 1/3 + ….+ (1/3)n
Notice, if and thusNotice, if and thus
the sum does not exist.the sum does not exist.
nSthenr 1
Looking at infinite series, what Looking at infinite series, what happens to the sum as n happens to the sum as n
approaches infinity if ? approaches infinity if ?
r
raS
n
n 1
11
So what if So what if ?1r
?1r
Sum of a Infinite Geometric Sum of a Infinite Geometric Series when Series when
r
aSn
1
1
n = # of termsn = # of terms
aa1 1 = 1= 1stst term term
r = common ratior = common ratio
?1r
Ex 5: Find the Sum of the infinite Ex 5: Find the Sum of the infinite seriesseries
32
1
9S
a) 1 + 1.5 + 2.25 + 3.375 + …a) 1 + 1.5 + 2.25 + 3.375 + …
Sum DNE since r = 1.5 and is > 1Sum DNE since r = 1.5 and is > 1
b) 9 + 6 + 4 + 8/3 + …b) 9 + 6 + 4 + 8/3 + …
r = 2/3 and is < 1 so we use the formular = 2/3 and is < 1 so we use the formula
27
319
H DubH Dub9-3 Pg.669 #3-42 (3n), 9-3 Pg.669 #3-42 (3n),
53-55, 73-75, 79-8153-55, 73-75, 79-81
