Geometric series and sequence ppts

1. 9-3 Geometric Sequences
& Series

2. 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).

3. 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

4. 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.

5. 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

6. 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

7. 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!

8. Formula for the Sum of a Finite
Geometric Series











r
r
a
S
n
n
1
1
1
n = # of terms
a1 = 1st term
r = common ratio

9. 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.











r
r
a
S
n
n
1
1
1























2
1
1
2
1
1
4
10
10
S
128
1023
1024
2046
4
2
1
1024
1023
4
2
1
1024
1
1
4
10 


































S























2
1
1
2
1
1
4
4
31
n

10. 






















2
1
1
2
1
1
4
4
31
n






















2
1
2
1
1
4
4
31
n
















n
2
1
1
8
4
31
n








2
1
1
32
31
n









2
1
32
1
n







2
1
32
1
5

n
n
n
2
1
32
1

n
2
32  log232=n

11. Looking at infinite series, what
happens to the sum as n
approaches infinity in each
case?
3 + 9 + 27 + 81, +….+ 3n
27 + 9 + 3, + 1 + 1/3 + ….+ (1/3)n
Notice, if and thus
the sum does not exist.


 n
S
then
r 1

12. Looking at infinite series, what
happens to the sum as n
approaches infinity if ?











r
r
a
S
n
n
1
1
1
So what if ?
1

r
?
1

r

13. Sum of a Infinite Geometric
Series when
r
a
Sn


1
1
n = # of terms
a1 = 1st term
r = common ratio
?
1

r

14. Ex 5: Find the Sum of the infinite
series









3
2
1
9
S
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
27
3
1
9









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