13.1, 13.3 Arithmetic and Geometric Sequences and Series13.5 Sums of Infinite Series

Objective

1. To identify an arithmetic or geometric sequence and find a formula for its n-th term;

2. To find the sum of the first n terms for arithmetic and geometric series.

1. Arithmetic Sequence and Series

1, 4, 7,10,13

9,1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

Arithmetic Sequences

ADDTo get next term

2, 4, 8,16, 32

9, 3,1, 1/ 3

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

, 2.5 , 6.25

Geometric Sequences

MULTIPLYTo get next term

Arithmetic Series

Sum of Terms

35

12

27.2

3 9

Geometric Series

Sum of Terms

62

20 / 3

85 / 64

9.75

a1 a2 a3 a4 a5 a6anan - 1

+ d + d + d + d + d

r

+ d

r r r r r

The notation of a sequence can be simply denoted as

{an}, meaning a1, a2, …, an, …

For any three consecutive terms in an arithmetic sequence,

… , ak – 1 , ak , ak + 1 , …

ak is called the arithmetic/geometric mean of ak – 1 and ak + 1 if the sequence is arithmetic/geometric.

If three consecutive terms,

… , ak – 1 , ak , ak + 1 , …

in an arithmetic sequence, then

ak = ak – 1 + dak = ak + 1 – d

ak + ak = ak – 1 + ak + 1

2ak = ak – 1 + ak + 1

Or, ak is the arithmetic average of ak – 1 and ak + 1.

k 1 k 1

k

a aa

2

If three consecutive terms,

… , ak – 1 , ak , ak + 1 , …

in a geometric sequence, then

Or, ak is the geometric average of ak – 1 and ak + 1.

k k k 1 k 1a a a a

k k 1

k 1 k

a ar

a a

2k k 1 k 1a a a

Geometric mean

k k k 1 k 1a a a a

2k k 1 k 1a a a

Arithmetic meanak + ak = ak – 1 + ak + 1

2ak = ak – 1 + ak + 1

Find the next four terms of –9, –2, 5, …

Arithmetic Sequence

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

7)2(5)9(2

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)

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

n 1

n 1 n

nth term of arithmetic sequence

sum of n terms of arithmetic sequen

a a n 1 d

nS a a

2ce

d

na n Sn 2

1Or 1

Given an arithmetic sequence with 15 1a 38 and d 3, find a .

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

x

15

38

NA

-3

n 1a a n 1 d

138 a 1 15 3

a1 = 80

63Find S of 19, 13, 7,...

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

-19

63

x

x

6

n 1a a n 1 d

63a 19 6 13 6 353

353

n 1 n

nS a a

2

63

633 3S

219 5

63 1 1S 052

15021]167[63]63119[6362

1631963Or 63

S

16 1Find a if a 1.5 and d 0.5 Try this one:

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

1.5

16

x

NA

0.5

n 1a a n 1 d

16 1.5 0.a 16 51

16a 9

n 1Find n if a 633, a 9, and d 24

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

9

x

633

NA

24

n 1a a n 1 d

633 9 n 1 24

633 9 24n 24

n = 27

1 29Find d if a 6 and a 20

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

-6

29

20

NA

x

n 1a a n 1 d

20 6 29 1 d

26 28d

13

d14

Find two arithmetic means between –4 and 5

-4, ____, ____, 5

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

-4

4

5

NA

x

n 1a a n 1 d

5 4 4 1 d

The two arithmetic means are –1 and 2, since –4, -1, 2, 5

forms an arithmetic sequence

d 3

Find three arithmetic means between 1 and 4

1, ____, ____, ____, 4

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

1

5

4

NA

x

n 1a a n 1 d

4 1 5 1 d

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

3

d4

Find n for the series in which 1 na 5, d 3, S 440

1a First term

na nth term

nS sum of n terms

n number of terms

d common difference

5

x

y

440

3

n 1a a n 1 d

n 1 n

nS a a

2

5 n 1 3

n

440 5 3n 22

n 7 3n440

2

880 n 7 3n

20 3n 7n 880

n = 16

( 16)(3 55) 0n n 55

16, or 3

n n 3n 2

Find the number of multiples of 4 between 35 and 225.

The difference of any two consecutive multiples of 4 is 4. Those multiples of 4 form an arithmetic sequence. Since 35 < 36 (the first multiple of 4 we are looking for), we can write the nth term of an arithmetic sequence as

36 ( 1) 4na n

35 225na

35 36 ( 1) 4 225n

1 4( 1) 189n

1 4 4 189n

3 4 193n

3 19348.25

4 4n

48n

1, 4, 7,10,13

9,1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

Arithmetic Sequences

ADDTo get next term

2, 4, 8,16, 32

9, 3,1, 1/ 3

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

, 2.5 , 6.25

Geometric Sequences

MULTIPLYTo get next term

Arithmetic Series

Sum of Terms

35

12

27.2

3 9

Geometric Series

Sum of Terms

62

20 / 3

85 / 64

9.75

2. Geometric Sequences and Series

a1 a2 a3 a4 a5 a6anan - 1

+ d + d + d + d + d

r

+ d

r r r r r

Vocabulary of Sequences (Universal)

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

n 1n 1

n

n 1

nth term of geometric sequence

sum of the first n terms of geometric seq

a a r

1 rue ce S a

1n

r

1 n

n

or

sum of the first n terms of geometric sequenca ra

S1

er

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

3 – 2 vs. 9/2 – 3… not arithmetic3 9 / 2 3

1.5 geometric r2 3 2

3 3 3 3 3 3

2 2 2

92, 3, , , ,

2

9 9 9

2 2 2 2 2 2

92, 3, , ,

27 81 243

4 8,

2 16

1 9

1 2If a , r , find a .

2 3

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

1/2

x

9

NA

2/3

n 1n 1a a r

9 1

9

1 2a

2 3

8

9 8

2a

2 3

7

8

2

3

128

6561

Find two geometric means between –2 and 54

-2, ____, ____, 54

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

-2

54

4

NA

x

n 1n 1a a r

4 154 2 r

327 r

r 3

The two geometric means are 6 and -18, since –2, 6, -18, 54

forms an geometric sequence

2 4 1

2Find a a if a 3 and r

3

-3, ____, ____, ____

2Since r ...

3

4 83, 2, ,

3 9

2 4

8 10a a 2

9 9

9Find a of 2, 2, 2 2,...

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

x

9

NA

2

2 2 2r 2

22

n 1n 1a a r

9

9

1

a 2 2

8

2 2

16 2

5 2 4If a 32 2 and r 2, find a to a .

____, , ____,________ ,32 2

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

x

5

NA

32 2

2n 1

n 1a a r

5

1

1

32 2 a 2

4

132 2 a 2

132 2 4a

1a 8 2

2a 8 2( 2) 16

23a 8 2( 2) 16 2

34a 8 2( 2) 32

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

*** denotes trick question

1,____,4

4

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

1/4

3

NA

4

xn 1

n 1a a r

3 114

4r 2r

14

4 216 r 4 r

1,1, 4

4

1, 1, 4

4

7

1 1 1Find S of ...

2 4 8

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

1/2

7

x

NA

11184r

1 1 22 4

n

n 1

1 rS a

1 r

7

7

11

1s

2 1

212

71

1 22

1

12

7

11

2

63

64

1, 4, 7, 10, 13, …. Infinite Arithmetic No Sum

3, 7, 11, …, 51 Finite Arithmetic

n 1 n

1

nS a a

2n 1

n a d2

1, 2, 4, …, 64 Finite Geometric

n1 n

n 1

a ra1 rS a

1 r 1 r

1, 2, 4, 8, … Infinite Geometricr 1 or r -1

No Sum

1 1 13,1, , , ...

3 9 27Infinite Geometric

-1 < r < 11a

S1 r

13.5 Infinite Geometric Series

|r| 1

|r| < 1

Find the sum, if possible: 1 1 1

1 ...2 4 8

1 112 4r

11 22

1 r 1 Yes

1a 1S 2

11 r 12

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

8 16 2r 2 2

82 2 r 1 No

NO SUM

Find the sum, if possible: 2 1 1 1

...3 3 6 12

1 113 6r

2 1 23 3

1 r 1 Yes

1

2a 43S

11 r 312

Find the sum, if possible: 2 4 8

...7 7 7

4 87 7r 22 47 7

r 1 No

NO SUM

Find the sum, if possible: 5

10 5 ...2

55 12r

10 5 2 1 r 1 Yes

1a 10S 20

11 r 12

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

32

32/5

40

32

32/5

40S 45

504

10

1554

0S 2 5 500 50

504

15

450

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

10S 80

100

4 43

1

0

10

3

An old grandfather clock is broken. When the pendulum is swung it follows a swing pattern of 25 cm, 20 cm, 16 cm, and so on until it comes to rest. What is the total distance the pendulum swings before coming to rest?

25

20

16

25

20

16

S

254

15

2 250

Find what values of x does the infinite series converge?

22 4 8x x ...

7 49 343

This infinite geometric series with r = 2x/7. By the theorem of convergence for the infinite geometric series, the series converges when | r | < 1. Or,

21

7

x

7

2x

7 7

2 2x or

Find what values of x does the infinite series converge?

23 91 (x 1) (x 1) ...

2 4

This infinite geometric series with r = 3(x – 1)/2. By the theorem of convergence for the infinite geometric series, the series converges when | r | < 1. Or,

3( 1) 1

2x

21

3x

2 21

3 3x or

1 5

3 3x So

Challenge Question

Sum of non-arithmetic and non-geometric series

Find the formula for the sum of the first n terms of the series, or Sn, then find the sum

1 1 1 13 5 7 (2 1)

3 9 27 3n nS n

The integer part of the each term forms a arithmetic series and the fraction part of each term forms a geometric series:

1 1 1 13 5 7 (2 1)

3 9 27 3n nS n

We apply two formula, one for arithmetic series and one for geometric series to get the overall formula for Sn:

1 11 1

1 1 3 33 2 ( 2)

12 3 213

n n

n

nS n n n

Challenge Question

Textbook P. 503 #30 (Sum of non-arithmetic and non-geometric series.)

Find the formula for the sum of the first n terms of the series, or Sn, then find the sum of the infinite series by using limit.

1 1 1 1

1 3 3 5 5 7 (2 1)(2 1)nSn n

Each term can be decomposed to half of the difference of two fractions – numerators are all 1 and the denominators are the two integers, respectively:

1 1 1 1 1 1 1 1 1

2 1 3 3 5 5 7 2 1 2 1nSn n

Starting from the second group, each first term within each parentheses can be cancel out by the second terms in the preceding group, respectively.

The only terms are not canceled out are 1 and –1/(2n + 1). So,

1 1 1 1 1 1 1 1 1

2 1 3 3 5 5 7 2 1 2 1nSn n

1 1 1 21

2 2 1 2 2 1 2 1n

n nS

n n n

Therefore the limit of the sum is

lim lim2 1n

n n

nS

n

lim

2 1n

nn

nn

1

lim1

2n

n

1

2

Given an arithmetic sequence {an}, how can you make a geometric sequence?

Challenge Question

We make a geometric sequence as . Since{ }nae

1k ka a d

1k ka a de e 1k ka a de e e

1

k

k

ad

a

ee

e

It shows that the new built sequence is a geometric sequence of common ratio r = ed .

Given a geometric sequence {an}, how can you make an arithmetic sequence?

Challenge Question

We make an arithmetic sequence as . Since{ln| |}na

1 1k k ka r a r a

It shows that the new built sequence is an arithmetic sequence of common difference d = ln|r| .

1ln lnk ka r a

1ln ln lnk ka r a

1ln ln lnk ka a r

B

nn A

a

UPPER BOUND(NUMBER)

LOWER BOUND(NUMBER)

SIGMA(SUM OF TERMS) NTH TERM

(SEQUENCE)

Index must be the same variable

Sigma Notation

The relationship between Sn , Sn – 1 and an

1 2 11

n

n n n kk

S a a a a a

1

1 1 2 11

n

n n kk

S a a a a

1n n nS S a

1n n na S S

1

1 1

n n

n k kk k

a a a

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

Arithmetic, d= 3

n 1a a n 1 d

na 3 n 1 3

na 3n4

1n

3n

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

Geometric, r = ½ n 1

n 1a a r n 1

n

1a 16

2

n 1

n

5

1

116

2

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

Not Arithmetic, Not Geometric

n 1na 20 2

n 1

n

5

1

20 2

19 + 18 + 16 + 12 + 4 -1 -2 -4 -8

Rewrite the following using sigma notation:3 9 27

...5 10 15

Numerator is geometric, r = 3Denominator is arithmetic d= 5

NUMERATOR: n 1

n3 9 27 ... a 3 3

DENOMINATOR: n n5 10 15 ... a 5 n 1 5 a 5n

SIGMA NOTATION: 1

1

n

n 5n

3 3

j

4

1

j 2

21 2 2 3 2 24 18

7

4a

2a 42 2 5 2 6 72 44

n

n 0

4

0.5 2

00.5 2 10.5 2 20.5 2 30.5 2 40.5 2

33.5

CAUTION

4

j 1

j (1 2 3 4)2 2

CAUTION

4 4 4

7

a a

7

a

7

( a) a a (4 5 6 7) 2 22 2 2 22 44

Some Properties of Sigma Notation

( )B B B

k k k kk A k A k A

a b a b

( 1)B

k A

c B A c

10

3k

m m m m

B B

k kk A k A

ca c a

1.

2.

3.

Ex.

15 15 152 2 2 2 2

5 5 5

( ) (5 6 15) (5 6 15 )k k k

k k k k

Ex.

8 times

(10 3 1) 8 (10 3)m m m

19 193 3 3 3 3

6 6

3 3 3(6 7 19 )k k

k k

Ex.

Some Useful Series

1

n

k

c nc

1

( 1)

2

n

k

n nk

1.

3. 4. 2

1

( 1)(2 1)

6

n

k

n n nk

23

1

( 1)

2

n

k

n nk

5.

1

1

1

nnk

k

aa a

a

2.

0

n

b

36

5

0

36

5

13

65

23

65

...

1aS

1 r

6

153

15

2

j

3

7

2j 1 2 1 2 8 1 2 9 1 ...7 2 123

n 1 n

2n 1S a a 15

2

3

2

747

527

2

j

3

7

2j 1

23 23

j j j7 7 7 1 1

23 23 6

j j

232j 1 2 j 7 1 1 2 j 2 j 17

1 123 23 6 62

( ) ( )2 2 17 24

26 7

23 17 527

1

b

9

4

4b 3

4 3 4 5 3 4 6 3 ...4 4 319

n 1 n

1n 1S a a 19

2

9

2

479

784

1

b

9

4

4b 3

19 19

b b b4 4 4 1 1

19 19 3

b b

194b 3 4 b 4 1 3 4 b 4 b 48

1 119 19 3 33

( ) ( )4 4 48 20

26 4

28 48 784

Assignment

13.1 P. 477 #4 – 36 (x4), 17 – 41(odd), 44 – 4534, 38, 47, 49, 51.

13.3 P. 489 #2 – 12 (even), 11, 15, 18 – 21, 23, 27, 30

13.5 P. 502 #2 – 18 (even), 21 – 35 (odd)