9.2 – Arithmetic Sequences and Series
An introduction…………
1, 4, 7, 10, 13
35
2, 4, 8, 16, 32
62
9, 1,  7,  15
12
9,  3, 1,  1/ 3
20 / 3
6.2, 6.6, 7, 7.4
27.2
,   3,   6
3  9
1, 1/ 4, 1/16, 1/ 64 85 / 64
9.75
, 2.5, 6.25
Arithmetic Sequences
Geometric Sequences
ADD
To get next term
MULTIPLY
To get next term
Arithmetic Series
Sum of Terms
Geometric Series
Sum of Terms
Find the next four terms of –9, -2, 5, …
Arithmetic Sequence
2  9  5  2  7
7 is referred to as the common difference (d)
Common Difference (d) – what we ADD to get next term
Next four terms……12, 19, 26, 33
Find the next four terms of 0, 7, 14, …
Arithmetic Sequence, d = 7
21, 28, 35, 42
Find the next four terms of x, 2x, 3x, …
Arithmetic Sequence, d = x
4x, 5x, 6x, 7x
Find the next four terms of 5k, -k, -7k, …
Arithmetic Sequence, d = -6k
-13k, -19k, -25k, -32k
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
d  common difference
nth term of arithmetic sequence  an  a1  n  1 d
sum of n terms of arithmetic sequence  Sn 
n
 a1  an 
2
Given an arithmetic sequence with a15  38 and d  3, find a1.
x
a1  First term
38 an  nth term
15
n  number of terms
NA Sn  sum of n terms
-3
d  common difference
an  a1  n  1 d
38  x  15  1 3 
X = 80
Find S63 of  19,  13, 7,...
-19 a1  First term
353
??
an  nth term
n  number of terms
63
x
Sn  sum of n terms
6
d  common difference
an  a1  n  1 d
??  19   63  1 6 
??  353
n
 a1  an 
2
63

 19  353 
2
Sn 
S63
S63  10521
Try this one: Find a16 if a1  1.5 and d  0.5
1.5 a1  First term
x
16
an  nth term
n  number of terms
NA Sn  sum of n terms
0.5
d  common difference
an  a1  n  1 d
a16  1.5  16  1 0.5
a16  9
Find n if an  633, a1  9, and d  24
9
a1  First term
633 an  nth term
x
n  number of terms
NA Sn  sum of n terms
24
d  common difference
an  a1  n  1 d
633  9   x  1 24
633  9  24x  24
X = 27
Find d if a1  6 and a29  20
-6
a1  First term
20 an  nth term
29
n  number of terms
NA Sn  sum of n terms
x
d  common difference
an  a1  n  1 d
20  6   29  1 x
26  28x
13
x
14
Find two arithmetic means between –4 and 5
-4, ____, ____, 5
-4
a1  First term
5
an  nth term
n  number of terms
4
NA
x
Sn  sum of n terms
d  common difference
an  a1  n  1 d
5  4   4  1 x 
x3
The two arithmetic means are –1 and 2, since –4, -1, 2, 5
forms an arithmetic sequence
Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
1
a1  First term
4
an  nth term
5
NA
x
n  number of terms
Sn  sum of n terms
d  common difference
an  a1  n  1 d
4  1   5  1 x 
3
x
4
The three arithmetic means are 7/4, 10/4, and 13/4
since 1, 7/4, 10/4, 13/4, 4 forms an arithmetic sequence
Find n for the series in which a1  5, d  3, Sn  440
5
a1  First term
y
an  nth term
x
n  number of terms
440 Sn  sum of n terms
3
d  common difference
an  a1  n  1 d
y  5   x  1 3
x
440   5  5   x  1 3 
2
x  7  3x 
440 
2
880  x  7  3x 
0  3x 2  7x  880
Graph on positive window
X = 16
n
Sn   a1  an 
2
x
440   5  y 
2
The sum of the first n terms of an infinite sequence
is called the nth partial sum.
Sn  n (a1  an)
2
Example 6. Find the 150th partial sum of the arithmetic sequence, 5,
16, 27, 38, 49, …
a1  5
d  11
 c  5  11  6
an  11n  6  a150  11150  6  1644
S150
150

 5  1644   75 1649   123,675
2
Example 7. An auditorium has 20 rows of seats. There are 20 seats in
the first row, 21 seats in the second row, 22 seats in the third row, and
so on. How many seats are there in all 20 rows?
d 1
c  20  1  19
an  a1   n 1 d  a20  20  19 1  39
20
S 20   20  39   10  59   590
2
Example 8. A small business sells $10,000 worth of sports memorabilia
during its first year. The owner of the business has set a goal of
increasing annual sales by $7500 each year for 19 years. Assuming that
the goal is met, find the total sales during the first 20 years this business
is in operation.
a1  10,000
d  7500
c  10,000  7500  2500
an  a1   n 1 d  a20  10,000  19  7500  152,500
20
S20  10,000  152,500   10 162,500   1,625,000
2
So the total sales for the first 2o years is $1,625,000
9.3 – Geometric Sequences and Series
1, 4, 7, 10, 13
35
2, 4, 8, 16, 32
62
9, 1,  7,  15
12
9,  3, 1,  1/ 3
20 / 3
6.2, 6.6, 7, 7.4
27.2
,   3,   6
3  9
1, 1/ 4, 1/16, 1/ 64 85 / 64
9.75
, 2.5, 6.25
Arithmetic Sequences
Geometric Sequences
ADD
To get next term
MULTIPLY
To get next term
Arithmetic Series
Sum of Terms
Geometric Series
Sum of Terms
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
r  common ratio
nth term of geometric sequence  an  a1r n1


a1 r n  1 

sum of n terms of geometric sequence  Sn  
r 1
Find the next three terms of 2, 3, 9/2, ___, ___, ___
3 – 2 vs. 9/2 – 3… not arithmetic
3 9/2
3

 1.5  geometric  r 
2
3
2
9 9 3 9 3 3 9 3 3 3
2, 3, ,  ,   ,   
2 2 2 2 2 2 2 2 2 2
9 27 81 243
2, 3, , , ,
2 4 8 16
1
2
If a1  , r  , find a9 .
2
3
a1  First term
1/2
an  nth term
x
n  number of terms
9
Sn  sum of n terms
NA
r  common ratio
2/3
an  a1r n1
 1  2 
x    
 2  3 
9 1
28
27
128
x
8 
8 
23
3
6561
Find two geometric means between –2 and 54
-2, ____, ____, 54
a1  First term
-2
an  nth term
54
n  number of terms
4
Sn  sum of n terms
NA
r  common ratio
x
an  a1r n1
54   2  x 
41
27  x 3
3  x
The two geometric means are 6 and -18, since –2, 6, -18, 54
forms an geometric sequence
Find a2  a 4 if a1  3 and r 
2
3
-3, ____, ____, ____
2
Since r  ...
3
3,  2,
4 8
,
3 9
 8  10
a 2  a 4  2  


9
 9 
Find a9 of 2, 2, 2 2,...
a1  First term
2
an  nth term
x
n  number of terms
Sn  sum of n terms
r  common ratio
9
NA
r
an  a1r n1
 2
2  2
x 2
x
x  16 2
9 1
8
2
2 2

 2
2
2
If a5  32 2 and r   2, find a 2
____, ____, ____,____,32 2
a1  First term
x
an  nth term
32 2
n  number of terms
Sn  sum of n terms
r  common ratio
an  a1r n1
NA
 2
 
2  x  2 
32 2  x  2
32
5
32 2  4x
8 2x
5 1
4
*** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1
,____,4
4
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
r  common ratio
an  a1r n1
1 2
1 31
4  r  4  r  16  r 2  4  r
4
4
1/4
4
3
NA
x
1
, 1, 4
4
1
,  1, 4
4
1 1 1
Find S7 of    ...
2 4 8
a1  First term
1/2
an  nth term
NA
n  number of terms
Sn  sum of n terms
7
x
r  common ratio


 a1 r n  1 

Sn  
r 1
 1   1 7

     1 
 2   2 
 
x

1
1
2
 1   1 7

     1 
 2   2 
  63

1
64

2
1 1
1
r 4  8 
1 1 2
2 4
Section 12.3 – Infinite Series
1, 4, 7, 10, 13, ….
Infinite Arithmetic
3, 7, 11, …, 51
Finite Arithmetic
1, 2, 4, …, 64
Finite Geometric
1, 2, 4, 8, …
Infinite Geometric
r>1
r < -1
No Sum
1 1 1
3,1, , , ...
3 9 27
Infinite Geometric
-1 < r < 1
a1
S
1 r
No Sum
n
Sn   a1  an 
2
a1 r n  1
Sn 
r 1
1 1 1
Find the sum, if possible: 1     ...
2 4 8
1 1
1
2
4
r  
 1  r  1  Yes
1 1 2
2
a1
S

1 r
1
1
1
2
2
Find the sum, if possible: 2 2  8  16 2  ...
8
16 2
r

 2 2  1  r  1  No
8
2 2
NO SUM
2 1 1 1
 ...
Find the sum, if possible:   
3 3 6 12
1 1
1
3
6
r  
 1  r  1  Yes
2 1 2
3 3
a1
S

1 r
2
3
4

1 3
1
2
2 4 8
   ...
Find the sum, if possible:
7 7 7
4 8
r  7  7  2  1  r  1  No
2 4
7 7
NO SUM
5
Find the sum, if possible: 10  5   ...
2
5
5
1
2
r
 
 1  r  1  Yes
10 5 2
a1
10
S

 20
1
1 r
1
2
The Bouncing Ball Problem – Version A
A ball is dropped from a height of 50 feet. It rebounds 4/5 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?
50
40
40
32
32
32/5
32/5
S
50
40

 450
4
4
1
1
5
5
The Bouncing Ball Problem – Version B
A ball is thrown 100 feet into the air. It rebounds 3/4 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?
100
100
75
75
225/4
225/4
S
100
100

 800
3
3
1
1
4
4
Sigma Notation
UPPER BOUND
(NUMBER)
B
SIGMA
(SUM OF TERMS)
a
n A
n
NTH TERM
(SEQUENCE)
LOWER BOUND
(NUMBER)
4
  j  2  1  2   2  2   3  2    4  2  18
j1
7
  2a    2  4    2 5    2  6   2 7  44
a4

4
n 0
 
 
 
 
 
4
3
2

0.5

2

0.5

2

0.5

2
0.5  2  0.5  2  0.5  2
n
 33.5
0
1


n
0
2
1
3
3
3
3
6

6

6

5
 5   6 
 5   ...

 
 
b 0 
5
a1
S

1 r
6
3
1
5
 15
  2x  1   2 7  1   2 8  1   2 9  1  ...   2  23  1
23
x 7
n
23  7  1
Sn   a1  an  
15  47   527
2
2
  4b  3    4  4  3   4 5  3   4  6  3   ...   4 19  3
19
b 4
Sn 
n
19  4  1
a

a

 1 n
19  79   784
2
2
Rewrite using sigma notation: 3 + 6 + 9 + 12
Arithmetic, d= 3
an  a1  n  1 d
an  3  n  1 3
an  3n
4
 3n
n1
Rewrite using sigma notation: 16 + 8 + 4 + 2 + 1
Geometric, r = ½
an  a1r n1
n1
 1
an  16  
2
 1
16  

2
n1
5
n1
Rewrite using sigma notation: 19 + 18 + 16 + 12 + 4
Not Arithmetic, Not Geometric
19 + 18 + 16 + 12 + 4
-1
-2 -4 -8
an  20  2n1
5
n1
20

2

n1
3 9 27

 ...
Rewrite the following using sigma notation: 
5 10 15
Numerator is geometric, r = 3
Denominator is arithmetic d= 5
NUMERATOR: 3  9  27  ...  an  3 3 
n1
DENOMINATOR: 5  10  15  ...  an  5  n  1 5  an  5n

SIGMA NOTATION:

n1
3 3 
n1
5n