Transcript 11.3 Geometric Sequences & Series Geometric Sequence • The ratio of a term to it’s previous term is constant. • This means you multiply.

11.3 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
Rule for a Geometric Sequence
n-1
an=a1r
Example: Write a rule for the nth term of the
sequence 5, 2, 0.8, 0.32,… . Then find a8.
•First, find r.
a8=5(.4)8-1
•r= 2/5 = .4
a8=5(.4)7
•an=5(.4)n-1
a8=5(.0016384)
a8=.008192
Example: One term of a geometric sequence
is a4=3. The common ratio is r=3. Write a rule
for the nth term. Then graph the sequence.
• If a4=3, then when
n=4, an=3.
• Use an=a1rn-1
3=a1(3)4-1
3=a1(3)3
3=a1(27)
1/ =a
9
1
• an=a1rn-1
an=(1/9)(3)n-1
• To graph, graph the
points of the form
(n,an).
• Such as, (1,1/9),
(2,1/3), (3,1), (4,3),…
Example: 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/ =a
If r=4, then a1=-1.
r
1
-1024=(-4/r)r5 an=(-1)(4)n-1
If r=-4, then a1=1.
4
-1024=-4r
n-1
a
=(1)(-4)
4
n
256=r
Both
n-1
a
=(-4)
4=r & -4=r
n
Work!
Formula for the Sum of a Finite
Geometric Series
1 r
S n  a1 
1

r

n
n = # of terms
a1 = 1st term
r = common ratio



Example: Consider the geometric
series 4+2+1+½+… .
• Find the sum of the
first 10 terms.
• Find n such that Sn=31/4.
n
1 r n 


S n  a1 
 1   1 
 1 r 
31   2 
 4
10


1
1
4
 1    
1



2 
2
S10  4

1 
 1



2


1   1023

1
 

2046 1023

1024
1024
  4
  4
S10  4

1

  1   1024 128

 

2

  2 







  1 n 
1   
31   2  
 4
1 
4
 1



2


31
1
 1  
32
 2
32  2
n
n
  1 n 
1   
31   2  
 4

1
4




2


1
1
  
32
2
n
log232=n
n

31   1  
 8 1  
 2 
4


1 1
 
32  2 
n5
n
1 1n
 n
32 2
Assignment