# 9.2 Arithmetic and Geometric Series and Sequences 9.2 – Arithmetic Sequences and Series

An introduction…………

1, 4, 7, 10, 13

9, 1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

35

12

27.2

3

 

9

Arithmetic Sequences

To get next term

Arithmetic Series

Sum of Terms

2, 4, 8, 16, 32

9,

3, 1, 1/ 3

1, 1/ 4, 1/16, 1/ 64

, 2.5 , 6.25

62

20 / 3

85 / 64

9.75

Geometric Sequences

MULTIPLY

To get next term

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) a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference nth term of arithmetic sequence

 a n

 a

1

    sum of n terms of arithmetic sequen ce

S n

 n

2

 a

1

 a n

Given an arithmetic sequence with a

15

38 and d

 

3, find a .

x

38

15

NA

-3 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

 a

1

   

38 x

1 5

1

  

X = 80

Find S

63 of

  

353

-19

??

63 x

6 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

 a

1

   

??

 

19

 

6 3

1

  

??

353

S n

S

63

 n

2

 a

1

 a n

63  

19

3 3

2

S

63

1 052 1

Try this one: Find a

16 if a

1

1.5 and d

0.5

1.5

x

16

NA

0.5

a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

 a

1

    a

16

1.5

a

16

9

 

16

1

0.

5

Find n if a n

633, a

1

9, and d

24

9

633 x

NA

24 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

 a

1

   

633 9

 x

1

2 4

633

 

2 4x

24

X = 27

Find d if a

1

 

6 and a

29

20 x

-6

20

29

NA a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

 a

1

   

20

 

6

 

29

1

 x

26

28x x

13

14

Find two arithmetic means between –4 and 5

-4

5

4

NA x

-4, ____, ____, 5 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a

5 n

 a

1

4

 

4

1

   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

5

NA x

1, ____, ____, ____, 4 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference a n

4

 a

1

1

5

1

 

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 a

1

5, d

3, S n

440

5 y x

440

3 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms d

 common difference

440

 x 

5

   x

1

3

2

440

 

3 x

2

880

 x 7

3x

0

3x

2 

7x

880

 a n

 a

1

    y 5

 x

1

3

Graph on positive window

X = 16

S n

 n

2

 a

1

 a n

4 40

 x

2

5

 y

sum

n

n

### th partial sum.

S n

 n

2

( a

1

 a n

)

Example 6. Find the 150 th partial sum of the arithmetic sequence, 5,

16, 27, 38, 49, … a

1

5 d

11 c 5 11 6 a n

11 n

6

 a

150

  

6 1644

S

150

    

### 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 a n a

1

 n 1

 d

 a

20

   

39

S

20

20 

20 39

    

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.

a

1

10,000 d

7500 c

10,000 7500

2500 a n 1

1

 d

 a

20

   

152,500

S

20

20 

10, 000 152,500

    

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

9, 1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

35

12

27.2

3

 

9

Arithmetic Sequences

To get next term

Arithmetic Series

Sum of Terms

2, 4, 8, 16, 32

9,

3, 1, 1/ 3

1, 1/ 4, 1/16, 1/ 64

, 2.5 , 6.25

62

20 / 3

85 / 64

9.75

Geometric Sequences

MULTIPLY

To get next term

Geometric Series

Sum of Terms

Vocabulary of Sequences (Universal) a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms r

 common ratio nth term of geometric sequence

 a n

 a r

1 sum of n terms of geometric sequ ence

S n

 a r

1

 n 

1

 r

1

Find the next three terms of 2, 3, 9/2, ___, ___, ___

3 – 2 vs. 9/2 – 3… not arithmetic

3

9 / 2

2 3

1.5

 geometric

 

3

2

9

2, 3, ,

2

9

2

3

2

,

9

2

3 3

2 2

,

9 3 3 3

2 2 2 2

9 27 81 243

2, 3, , ,

2 4 8

,

16

If a

1

1

, r

2

2 3

, find a .

9 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms r

 common ratio

1/2 x

9

NA

2/3 a n

 a r

1

1 2 x

9

1 x

2 8

8

2 7

3

8

128

6561

Find two geometric means between –2 and 54

-2, ____, ____, 54 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms r

 common ratio

-2

54

4

NA x a n

 a r

54

   

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 a

2

 a if a

4 1

 

3 and r

2

3

-3, ____, ____, ____

Since r

2

3

...

 

4

3, 2, ,

3 9

8 a

2

 a

4

2

 

8

 9 

10

9

Find a of 2, 2, 2 2,...

9 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms r

 common ratio x

9

2

NA r

2

2 2

2 2

 a n

 a r x x

2

2

 

 

9

1

8 x

16 2

2

If a

5

32 2 and r

 

2, find a

2

____, ____ , ____,____ ,32 2 a

1

First term a n

 nth term n

 number of terms x

32 2

5

S n

 sum of n terms r

 common ratio

NA

 a n

 a r

32 2

 x

 

5

1

32 2

 x

 

4

2

32 2

4 x

8 2

 x

4

*** Insert one geometric mean between ¼ and 4***

*** denotes trick question

1

4 r 3

1

1

,____,4

4 a

1

First term a n

 nth term n

 number of terms

S n

 sum of n terms r

 common ratio a n

 a r

1

1 r

2

4

16

 r

2  

4

 r

1/4

4

3

NA x

1

4

1

4

, 1, 4

Find S of

7

1 1 1

2 4 8

...

S n

 a

1

First term a n

 nth term n

 number of terms a r

1

 n 

1

 r

1

S n

 sum of n terms r

 common ratio x

1

1

2 2

1

1

2

1

1

1

2 2

1

2

1/2

NA

7

1

63

64 x r

1 1

 4  8 

1

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

1, 2, 4, 8, …

Finite Geometric

Infinite Geometric r > 1 r < -1

1 1 1

3,1, , , ...

3 9 27

Infinite Geometric

-1 < r < 1

No Sum

S n

S n

 n  a

1

 a n

2 a r

1

 n 

1

 r

1

No Sum

S

 a

1

Find the sum, if possible: 1

1 1 1

2 4 8

1 1 r

 2  4 

1 1

2

1

2

1 r 1

...

Yes

S

 a

1

1

1

1

2

2

Find the sum, if possible: 2 2 8 16 2

...

r

8

16 2

2 2 8

2 2 1 r 1 No

NO SUM

Find the sum, if possible:

2 1 1 1

3 3 6 12

...

r

1 1

 3  6 

1

2 1 2

3 3

1 r

S

 a

1

2

1

3

1

2

4

3

1 Yes

Find the sum, if possible:

2 4 8

7 7 7

...

4 8 r

 7  7 

2 4

2

7 7

1 r 1 No

NO SUM

Find the sum, if possible: 10 5

2

...

r

5

5

 2 

1

10 5 2

1 r 1 Yes

S

 a

1

10

1

1

2

20

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

1

4

5

40

1

4

5

45 0

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

75

225/4

100

75

225/4

S

100

1

3

4

10

1

0

3

4

80 0

Sigma Notation

UPPER BOUND

(NUMBER)

SIGMA

(SUM OF TERMS)

B  a n NTH TERM

(SEQUENCE)

LOWER BOUND

(NUMBER)

4  j

1

 j

2

 

2

  

2

2

  

3

2

 

4

2

 

18

7  a

4

2

 

 

2

44

4  n

0

0.5

2 n

0.5

2 0

33.5

0.5

2 1

0.5

2 2

0.5

2 3

0.5

2 4

 b

0

6

3

 

6

3

 

0

S

 a

1

3

1

6

6

1

3

5

15

2

...

x

2 3 

7

2x

1

S n

2 1

 n

2

 a

1

 a n

 

2 3

 

2

   

1 

15

47

527

2

1

1 9 b

4

4b

3

 

4

3

S n

 n

2

 a

1

 a n

 

1 9

 

2

3

  

1 

19

79

3

 

784

4

3

Rewrite using sigma notation: 3 + 6 + 9 + 12

Arithmetic, d= 3 a n

 a

1

    a n

     a n

3n

4  n

1

3n

Rewrite using sigma notation: 16 + 8 + 4 + 2 + 1

Geometric, r = ½ a n

 a r a n

16

1

 

5  n

1

16

1

 

Rewrite using sigma notation: 19 + 18 + 16 + 12 + 4

Not Arithmetic, Not Geometric

19 + 18 + 16 + 12 + 4

-1 -2 -4 -8 a n

20

2

5  n

1

20

2

Rewrite the following using sigma notation:

3

9

27

5 10 15

...

Numerator is geometric, r = 3

Denominator is arithmetic d= 5

NUMERATOR: 3

    a n

  

DENOMINATOR:

    a n

      a n

5n

SIGMA NOTATION: n

1

  n

1

5n