arithmetic sequences

Arithmetic Sequences

1

USING AND WRITING SEQUENCES

The numbers in sequences are called terms.

You can think of a sequence as a function whose domain is a set of consecutive integers. If a domain is not specified, it is understood that the domain starts with 1.

2

The domain gives the relative position of each term.

1 2 3 4 5 DOMAIN:

3 6 9 12 15RANGE:The range gives the terms of the sequence.

This is a finite sequence having the rulean = 3n,

where an represents the nth term of the sequence.

USING AND WRITING SEQUENCES

n

an

3

Writing Terms of Sequences

Write the first six terms of the sequence an = 2n + 3.

SOLUTION

a 1 = 2(1) + 3 = 5 1st term

2nd term

3rd term

4th term

6th term

a 2 = 2(2) + 3 = 7

a 3 = 2(3) + 3 = 9

a 4 = 2(4) + 3 = 11

a 5 = 2(5) + 3 = 13

a 6 = 2(6) + 3 = 15

5th term

4

Writing Terms of Sequences

Write the first six terms of the sequence f (n) = (–2) n – 1 .

SOLUTION

f (1) = (–2) 1 – 1 = 1 1st term

2nd term

3rd term

4th term

6th term

f (2) = (–2) 2 – 1 = –2

f (3) = (–2) 3 – 1 = 4

f (4) = (–2) 4 – 1 = – 8

f (5) = (–2) 5 – 1 = 16

f (6) = (–2) 6 – 1 = – 32

5th term

5

An introduction…………

1, 4, 7,10,139,1, 7, 156.2, 6.6, 7, 7.4, 3, 6

ARITHMETIC

ADD(by the same #)

To get the next term

2, 4, 8,16, 329, 3,1, 1/ 31,1/ 4,1/16,1/ 64

, 2.5 , 6.25

GEOMETRIC

MULTIPLY(by the same #)

To get the next term

d = 3 d = -8 d = .4 d = 3

r =2

r = 41

5.2r =

6

r = 31

Vocabulary of Sequences (Universal)

1a First term

na nth term

n number of termsd common difference

r common ratio

Finite VS. Infinite 7

an-1 previous term an+1 next term

Arithmetic Sequence: sequence whose consecutive terms have a common difference.

Example: 3, 5, 7, 9, 11, 13, ...

The terms have a common difference of 2. (known as d)

To find the common difference you use an+1 – an Example: Is the sequence arithmetic? If so, find d.

–45, –30, –15, 0, 15, 30 d = 15

8

Find the next 4 terms of –9, -2, 5, …

2 9 5 2 7 7 is referred to as d

Next four terms…… 12, 19, 26, 33

9

Arithmetic Sequence, d = 7 21, 28, 35, 42

Arithmetic Sequence, d = x 4x, 5x, 6x, 7x

Find the next four terms of 0, 7, 14, …

Find the next four terms of x, 2x, 3x, …

Find the next four terms of 5k, -k, -7k, …

Arithmetic Sequence, d = -6k

-13k, -19k, -25k, -31k10

The nth term of an arithmetic sequence is given by:

1 ( 1)na a n d The nth term in the sequence

First term

The common difference

The term #

)6(346664)6(24664)6(1464)6(044

4

3

2

1

aaaa

4, 10, 16, 22

585446)110(410 aFind the 10th term:

11

Find the 14th term of the sequence: 4, 7, 10, 13,

……1 ( 1)na a n d

14a 4 (13)3 43

3)114(4

12

1 ( 1)na a n d

In the arithmetic sequence 4,7,10,13,…, which term has a value of 301?

301 4 ( 1)3n 301 4 3 3n 301 1 3n 300 3n 100 n

13

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

n 1a a n 1 d

38 x 1 15 3

X = 8014

1 29Find d if a 6 and a 20

120 6 29 x

26 28x13x14

15

Example: If the common difference is 4 and the fifth term is 15, what is the 10th term of an arithmetic sequence?

an = a1 + (n – 1)d d = 4, a5 = 15, n = 5, a1=?

15 = a1 + (5 – 1)4 15 = a1 +16 a1 = –1

a10 = –1 + (10 – 1)4= -1 + 36

a10 = 35 16

Explicit Formula – used to find the nth term of the arithmetic sequence in which the common difference and 1st term are known.

Ex: 4, 6, 8, 10…

Use a1 and d in sequence formula: an = 4 + (n – 1)2 an = 2n + 2

17

Explicit vs. Recursive Formulas

Find the explicit formula for the following arithmetic sequence:3, 8, 13, 18…

an = a1 + (n – 1)d a1 = 3 d = 5 n = ?

an = 3 + (n – 1)5 an = 3 + 5n – 5

an = -2 + 5n OR an = 5n – 2 18

Recursive Formula – (includes a1) used to find the next term of the sequence by adding the common difference to the previous term.

19

Explicit vs. Recursive Formulas

an = an-1 + 2 a1 = 4Ex: 4, 6, 8, 10…

an = an-1 + da1 = ___

an+1 = an + d a1 = ___

Find the recursive formula for the following arithmetic sequence:3, 8, 13, 18…

an = an-1 + d a1 = 3 d = 5

an = an-1 + 5 a1 = 3

20

21

Using Recursive & Explicit Formulas

an = an-1 + 6 a1 = 4

1. Create the 1st 5 terms:4, 10, 16, 22, 28

2. Find the explicit formula:

an = a1 + (n – 1)dan = 4 + (n – 1)6an = 4 + 6n – 6 an = 6n – 2

a2 = 4 + 6 = 10 a3 = 10 + 6 = 16 a4 = 16 + 6 = 22 a5 = 22 + 6 = 28

22

Using Recursive & Explicit Formulas

an = an-1 – 2 a1 = 5

1. Create the 1st 5 terms:5, 3, 1, –1, –3

2. Find the recursive formula:

an = 7 – 2n

a2 = 7 – 2(2) = 3

a5 = 7 – 2(5) = –3 a4 = 7 – 2(4) = –1 a3 = 7 – 2(3) = 1

a1 = 7 – 2(1) = 5

An arithmetic mean of two numbers, a and b, is simply their average. Use the formula and information given to find the common difference to create the sequence. Insert 3 arithmetic

means between 8 & 16.

16 8 (5 1)d 2d

Let 8 be the 1st termLet 16 be the 5th termLet 5 be Nd is missing

1 ( 1)na a n d

8, , , ,1610 12 14

23

Find two arithmetic means between –4 and 5 -4, ____, ____, 5

n 1a a n 1 d 15 4 4 x

x 3The two arithmetic means are –1 and 2,

since –4, -1, 2, 5 forms an arithmetic sequence24

Find 3 arithmetic means between 1 & 41, ____, ____, ____, 4

n 1a a n 1 d 4 1 x15

3x4

The 3 arithmetic means are

since 1, ,4 forms a sequence4

13,4

10,47

413,

410,

47

25

Geometric Sequences

26

Vocabulary of Sequences (Universal)

1a First term

na nth term

n number of termsd common difference

r common ratio

an-1 previous term an+1 next term

27Finite VS. Infinite

Find the next 3 terms of 2, 3, 9/2, __, __, __

3 – 2 vs. 9/2 – 3… not arithmetic

3 9 / 2 31.5 geometric r2 3 2

• Use to determine common ration

n

aa 1

28

23

23

232

4th term:

29

n 1n 1a a r

The nth term of a geometric sequence is given

by:

23

23

23

232

5th term:

23

23

23

23

232

6th term:

1st term: 2

3232 : term2nd

29

23

232 : term3rd

How is the formula derived?

1 91 2If a , r , find a .2 3

n 1n 1a a r

9 11 2x2 3

8

8

2x2 3

7

8

23

1286561

30

2 4 12Find a a if a 3 and r3

2Since r ...3

2 48 10a a 2

9 9

31

-3, ____, ____, ____2 34

98

9Find a of 2, 2, 2 2,...n 1

n 1a a r 9 1x 2 2

8x 2 2

x 16 232

r = 2a1= 2

n = 9

5 2If a 32 2 and r 2, find a ____, , ____,________ ,32 2

n 1n 1a a r

5 132 2 x 2

432 2 x 2

32 2 x4

8 2 x

21 2282 a

33

1648

2281

Ex: 4, 12, 36, 108…

Use a1 and r in sequence formula:

34

Explicit vs. Recursive FormulasExplicit Formula – used to find the nth term of the geometric sequence in which the common ratio and 1st term are known.

Ex: an = a1*rn-1 an = 4 * 3n-1

Find the explicit formula for the following geometric sequence:3, 6, 12, 24…

an = a1*rn-1 a1 = 3 r =2

an = 3 *2n-1

35

36

Explicit vs. Recursive Formulas

an = an-1 (–4) a1 = –1

Ex: –1, 4, –16, 64 …

an = an-1 (r)a1 = ___

an+1 = r(an) a1 = ___

Recursive Formula (includes a1) – used to find the next term of the sequence by multiplying the common ratio to the previous term.

a1 (r) = a2 a2 (r) = a3a3 (r) = a4

Find the recursive formula for the following geometric sequence:3, 6, 12, 24…an = an-1 * r a1 = 3 r = 2

an = an-1 * 2 a1 = 3

37

38

Using Recursive & Explicit Formulas

an = an-1 (3) a1 = –1

1. Create the 1st 5 terms:–1, –3, –9, –27, – 81

2. Find the explicit formula:

an = a1 (r)n-1

an = –1(3)n-1

a2 = –1(3) = –3a3 = –3(3) = –9 a4 = –9(3) = –27 a5 = –27(3) = –81

an = –3n-1

39

Using Recursive & Explicit Formulas

an = 4an-1

a1 = 2

1. Create the 1st 5 terms:2, 8, 32, 128, 5122. Find the recursive formula:

an = 2(4)n – 1

a2 = 2(4)2-1 = 8

a5 = 2(4)5-1 = 512a4 = 2(4)4-1 = 128a3 = 2(4)3-1 = 32

a1 = 2(4)1-1 = 2

Ex: Find two geometric means between –2 and 54

-2, ____, ____, 54

n 1n 1a a r 1454 2 x

327 x 3 x

40

A geometric mean(s) of numbers are the terms between any 2 nonsuccessive terms of a geometric sequence. Use the terms given to find the common ratio and find the missing terms called the geometric means.

The 2 geometric means are 6 and -18

6 –18

*** Insert one geometric mean between ¼ and 4****** denotes trick question

1,____,44

n 1n 1a a r

3 1144

r 2r144

216 r 4 r 1,1, 44

1, 1, 44

41

Series42

Vocabulary of Sequences (Universal)

1a First term

na nth term

n number of termsd common difference

r common ratio

an-1 previous term an+1 next term

43Finite VS. Infinite

USING SERIES

. . .

FINITE SEQUENCE

FINITE SERIES3, 6, 9, 12, 15

3 + 6 + 9 + 12 + 15

INFINITE SEQUENCE

INFINITE SERIES

3, 6, 9, 12, 15, . . .

3 + 6 + 9 + 12 + 15 + . . .

When the terms of a sequence are added, the resulting expression is a series. A series can be finite or infinite.

44

You can use summation notation to write a series. For example, for the finite series shown above, you can write

3 + 6 + 9 + 12 + 15 = ∑ 3i5

i = 1

B

nn A

a

UPPER BOUNDTERM NUMBER

LOWER BOUNDTERM NUMBER

SIGMA(SUM OF TERMS)

NTH TERMSEQUENCE

(EXPLICIT FORMULA)

45

# of Terms: B – A + 1

j

4

1

j 2

21 2 2 3 2 24

18

7

4a

2a

42 2 5 2 6 72

4446

An arithmetic series is a series associated

with an arithmetic sequence.

It can be infinite or finite.

47

1, 4, 7, 10, 13, ….Infinite Arithmetic

(constantly getting larger or smaller)

3, 7, 11, …, 51 Finite Arithmetic n 1 nnS a a2

1, 2, 4, …, 64

1, 2, 4, 8, …

1 1 13,1, , , ...3 9 27

No Sum

48

Find the sum of the 1st 100 natural numbers.

1 + 2 + 3 + 4 + … + 100

12n nnS a a

1 1a 100na 100n 100

100 (1 100)2

S

505049

Find the sum of the 1st

14 terms of the series: 2 + 5 + 8 + 11 + 14 + 17 +…

a14 = 2 + (14 - 1)(3) = 41

301 S14 = 4122

14

50

1414 22

14 aS

1 2a 14n nn aanS 12

To find a14

, you need 3d da )114(214

13

1

(4 5)n

n

Find the sum of the series

9 13 17 ....

1 9a 4d 13n

13(66)

2 429

12n nnS a a

1313 92

13 aS

Need 13th term:

4(13) + 5 = 57

51

5792

1313 S

n = 4 a1 = 3 a4 = 6

12n nnS a a

18)9(26324

4 S

j

4

1

j 2

7

4a

2a

Finding the Sum from Summation Notation

n = (7 – 4) + 1 a4 = 8 a7 = 14

44)22(214824

4 S52

3, 4, 5, 6

8, 10, 12, 14

527

2

x

3

7

2x 1

1

b

9

4

4b 3

784a4 =19 a19 = 79 n = (19 - 4) + 1 = 16

)98(8)7919(2

1616 S

)62(5.8)4715(2

1717 S

a7 =15 a23 = 47 n = (23-7) + 1 = 17

53

19, 23, 27, 31…79

15, 17, 19, …47

An geometric series is a series associated with a geometric sequence. They can be

infinite or finite. Finite and infinite have

different formulas depending on the

value of r.54

1, 4, 7, 10, 13, …. Infinite Arithmetic(constantly getting larger or smaller)

3, 7, 11, …, 51 Finite Arithmetic n 1 nnS a a2

1, 2, 4, …, 64 Finite Geometric

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

(constantly getting larger or smaller)

“diverges”

Infinite Geometric-1 < r < 1

“converges”

No Sum

No Sum

rraS

n

n

1

)1(1

55r

aS

1

1...

271,

91,

31,1,3

71 1 1Find S of ...2 4 8

11184r

1 1 22 4

?

721

1)1(

71:

11

rn

arraS

termsstofsumFindFiniten

n

211

))21(1(

21 7

7

S

128127

56

Sums of Infinite Series Made Finite

(referred to as partial sums)

Infinite SeriesFinding the Sum

of Infinite Sequences“Converges” vs. “Diverges”

57

58

Find the sum, if possible:

1 1 11 ...2 4 8

1 1

12 4r11 22

Is -1 < r < 1? Yes (Infinite Series - converges)

59

raS

1

1

211

1

S 2

211

Geometric~need to find r~

Find the sum, if possible:

2 2 8 16 2 ... 8 16 2r 2 2

82 2

NO SUM Is -1 < r < 1? No (Infinite series - Diverges)

60

Find the sum, if possible:

Is -1 < r < 1? Yes (Infinite Series – Converges)

61

raS

1

1

321

1

S 3

...278

94

321

32

3294

132

r

Find the sum, if possible:

2 4 8 ...7 7 7

4 87 7r 22 47 7

NO SUM

Is -1 < r < 1? No (Infinite Series–Diverges)

62

Find the sum, if possible: 510 5 ...2

5

5 12r10 5 2

-1 < r < 1 Yes (Infinite Series–Converges)

63

raS

1

1

211

10

S 20

0

n

b

365

036

5

1365

2365

...

1aS1 r

6 15315

2

x

3

7

2x 1

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

n 1 n2n 1S a a 15

23

27 47

527

47...,19,17,15

64

Finding the Sum from Sigma Notation

53

r so “converges”

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

Arithmetic, d= 3

n 1a a n 1 d

na 3 n 1 3

Explicit formula

65

4th term

4

1st term

n=1

na 3n n3

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

Geometric, r = ½ n 1

n 1a a r

66

Explicit formula

n 1

n1a 162

1

2116

n

n=1

1st term

5

5th term

Rewrite the following using sigma notation:3 9 27 ...5 10 15

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

NUMERATOR: n 1n3 9 27 ... a 3 3

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

SIGMA NOTATION: 1

1

n

n 5n3 3

67