Sequence And Series

Copyright © 2007 Pearson Education, Inc. Slide 11-1

DONE BY,

AFSAL M NAHAS XI B KV PATTOM


SEQUENCE AND SERIES







An introduction…………

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


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)

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


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

38 x 1 15 3

X = 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

6

n 1a a n 1 d

?? 19 6 1

?? 353

3 6

353

n 1 n

nS a a

2

63

633 3S

219 5

63 1 1S 052


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 21x 4

633 9 2 244x

X = 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

120 6 29 x

26 28x

13x

14


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

15 4 4 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

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 x15 3

x4

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 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

y 5 31x

x40 y4

25

12

x440 5 5 x 3

x 7 x440

2

3

880 x 7 3x 20 3x 7x 880

X = 16

Graph on positive window


Copyright © 2007 Pearson Education, Inc. Slide 11-21Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 21

An infinite sequence is a function whose domain is the set of positive integers.

a1, a2, a3, a4, . . . , an, . . .

The first three terms of the sequence an = 2n2 are

a1 = 2(1)2 = 2

a2 = 2(2)2 = 8

a3 = 2(3)2 = 18.

finite sequence

terms


A sequence is geometric if the ratios of consecutive terms are the same.

2, 8, 32, 128, 512, . . .

geometric sequence

The common ratio, r, is 4.

82

4

328

4

12832

4

512128

4


The nth term of a geometric sequence has the form

an = a1rn - 1

where r is the common ratio of consecutive terms of the sequence.

15, 75, 375, 1875, . . . a1 = 15

The nth term is 15(5n-1).

75 515

r

a2 = 15(5)

a3 = 15(52)

a4 = 15(53)


Example: Find the 9th term of the geometric sequence

7, 21, 63, . . .

a1 = 7

The 9th term is 45,927.

21 37

r

an = a1rn – 1 = 7(3)n – 1

a9 = 7(3)9 – 1 = 7(3)8

= 7(6561) = 45,927


The sum of the first n terms of a sequence is represented by summation notation.

1 2 3 41

n

i ni

a a a a a a

index of summation

upper limit of summation

lower limit of summation

5

1

4n

n

1 2 3 4 54 4 4 4 4 4 16 64 256 1024 1364


The sum of a finite geometric sequence is given by

11 1

1

1 .1

n nin

i

rS a r ar

5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ?

n = 8

a1 = 5

1

81 11

221

5n

nrS ar

5210r

1 25651 2 2555

1 1275


The sum of the terms of an infinite geometric sequence is called a geometric series.

a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . .

If |r| < 1, then the infinite geometric series

11

0

.1

i

i

aS a r

r

has the sum

If 1 , then the series does not have a sum.r


Example: Find the sum of

1

1a

Sr

1 13 13 9

13

r

3

1 13

3 31 413 3

The sum of the series is 9 .4

3 934 4


11.3 Geometric Sequences and Series

1, 2, 4, 8, 16 … is an example of a geometric sequence with first term 1 and each subsequent term is 2 times the term preceding it.

The multiplier from each term to the next is called the common ratio and is usually denoted by r.

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant nonzero real number.


11.3 Finding the Common Ratio

In a geometric sequence, the common ratio can be found by dividing any term by the term preceding it.

The geometric sequence 2, 8, 32, 128, …has common ratio r = 4 since

8 32 128... 4

2 8 32



nth Term of a Geometric Sequence

In the geometric sequence with first term a1 and common ratio r, the nth term an, is

11

nna a r


11.3 Using the Formula for the nth Term

Example Find a5 and an for the geometric

sequence 4, –12, 36, –108 , …

Solution Here a1= 4 and r = 36/ –12 = – 3. Using

n=5 in the formula

In general

5 1 45 4 ( 3) 4 ( 3) 324a

1 11 4 ( 3)n n

na a r

11

nna a r


11.3 Modeling a Population of Fruit Flies

Example A population of fruit flies grows in such a way that each generation is 1.5 times the previous generation. There were 100 insects in the first generation. How many are in the fourth generation.

Solution The populations form a geometric sequence with a1= 100 and r = 1.5 . Using n=4 in the formula

for an gives

or about 338 insects in the fourth generation.

3 34 1 100(1.5) 337.5a a r


11.3 Geometric Series

A geometric series is the sum of the terms of a geometric sequence .

In the fruit fly population model with a1 = 100 and r = 1.5, the total population after four generations is a geometric series:

1 2 3 4

2 3100 100(1.5) 100(1.5) 100(1.5)

813

a a a a



Sum of the First n Terms of an Geometric Sequence

If a geometric sequence has first term a1 and common ratio r, then the sum of the first n terms is given by

where .1(1 )

1

n

n

a rS

r

1r


11.3 Finding the Sum of the First n Terms

Example Find

Solution This is the sum of the first six terms of a

geometric series with and r = 3.

From the formula for Sn ,

.

11 2 3 6a

6

1

2 3i

i

6

6

6(1 3 ) 6(1 729) 6( 728)2184

1 3 2 2S


Vocabulary of Sequences (Universal)

1a First term

na nth term

nS sum of n terms

n number of terms

r common ratio

n 1

n 1

n1

n

nth term of geometric sequence

sum of n terms of geometric sequ

a a r

a r 1S

r 1ence


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 11 2

x2 3

8

8

2x

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

1454 2 x

327 x 3 x

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 1

x 2 2

8

x 2 2

x 16 2


5 2If a 32 2 and r 2, find 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

32 2 x 2

4

32 2 x 2

32 2 x4

8 2 x


*** 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

n1

n

a r 1S

r 1

71 12 2

x12

1

1

71 12 2

12

1

63

64








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

3, 7, 11, …, 51 Finite Arithmetic n 1 n

nS a a

2

1, 2, 4, …, 64 Finite Geometric n

1

n

a r 1S

r 1

1, 2, 4, 8, … Infinite Geometricr > 1r < -1

No Sum

1 1 13,1, , , ...

3 9 27Infinite Geometric

-1 < r < 11a

S1 r


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 1 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

1 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


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


11.3 Infinite Geometric Series

If a1, a2, a3, … is a geometric sequence and the sequence of sums S1, S2, S3, …is a convergent

sequence, converging to a number S. Then S is

said to be the sum of the infinite geometric series 1 2 3 ...a a a S


11.3 An Infinite Geometric Series

Given the infinite geometric sequence

the sequence of sums is S1 = 2, S2 = 3, S3 = 3.5, …

1 1 1 12, 1, , , , ,...

2 4 8 16

The calculator screen shows more sums, approaching a value of 4. So

1 12 1 ... 4

2 4


11.3 Infinite Geometric Series

Sum of the Terms of an Infinite Geometric Sequence

The sum of the terms of an infinite geometric sequence with first term a1 and common ratio r, where –1 < r < 1 is given by

.1

1

aS

r


11.3 Finding Sums of the Terms of Infinite Geometric Sequences

Example Find

Solution Here and so

.

1

3

5

i

i

1

3

5a

1

1

33 35

35 1 215

i

i

a

r

3

5r


Sequence And Series

Education

terms of x

copyright bypearson

x n number of terms

arithmetic sequence

x d common difference30

sn sum of n terms880

arithmetic means

term 5x440