10.2 – Arithmetic Sequences and Series

An introduction … describe the pattern

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

− −

π π + π +

Arithmetic SequencesADD

To get next term

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

, 2.5 , 6.25

− −

π π π

Geometric SequencesMULTIPLY

To get next term

Arithmetic SeriesSum of Terms

3512

27.23 9

−

π +

Geometric SeriesSum of Terms

6220 / 385 / 649.75π

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 = 721, 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 dnS a a2

ce

= + −

= +

→

→

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

1538

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 nnS a a2

= +

( )6363 3 3S2

19 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

16x

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

x633

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

2920

NA

x

( )n 1a a n 1 d= + −

( )120 6 29 x= + −−

26 28x=

13x14

=

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

45

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 41, ____, ____, ____, 4

1a First term→

na nth term→

nS sum of n terms→

n number of terms→

d common difference→

1

54

NA

x( )n 1a a n 1 d= + −

( )( )4 1 x15= + −3x4

=

The three arithmetic means are 7/4, 10/4, and 13/4since 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

xy

440

3

( )n 1a a n 1 d= + −

( )n 1 nnS a a2

= +

( )y 5 31x= + −

( )x40 y42

5= +

( )( )12x440 5 5 x 3= + + −

( )x 7 x440

23

=+

( )880 x 7 3x= +20 3x 7x 880= + −

X = 16

Graph on positive window

10.3 –

Geometric Sequences and Series

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

− −

π π + π +Arithmetic Sequences

ADDTo get next term

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

, 2.5 , 6.25

− −

π π π

Geometric SequencesMULTIPLY

To get next term

Arithmetic SeriesSum of Terms

3512

27.23 9

−

π +

Geometric SeriesSum of Terms

6220 / 385 / 649.75π

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 arithmetic

3 9 / 2 31.5 geometric r2 3 2= = → → =

92, 3, , ,27 81 2434 8

,2 16

To find r, divide any term in the sequence by its preceding term.

a2 /a1 a3 /a2

1 91 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→

1/2x

9NA2/3

n 1n 1a a r −=

9 11 2x2 3

−⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

8

82x

2 3=

×

7

823

= 1286561

=

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→

-254

4NAx

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

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

9NA

2

2 2 2r 222

= = =

n 1n 1a a r −=

( )9 1x 2 2

−=

( )8x 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

5NA

32 2

2−n 1

n 1a a r −=

( )5 132 2 x 2

−−=

( )432 2 x 2= −

32 2 x4=

8 2 x=

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

1,____,44

1a First term→

na nth term→

nS sum of n terms→

n number of terms→

r common ratio→

1/4

3NA

4

xn 1

n 1a a r −=

3 1144

r −= 2r144

→ = 216 r→ = 4 r→ ± =

1,1, 44

1, 1, 44

−

71 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

7x

NA

11184r

1 1 22 4

= = =

( )n1

n

a r 1S

r 1

⎡ ⎤−⎣ ⎦=−

71 12 2

x12

1

1

⎡ ⎤⎛ ⎞⎛ ⎞ −⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦=−

71 12 2

12

1⎡ ⎤⎛ ⎞⎛ ⎞ −⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦=

−

6364

=

Section 12.3 – Infinite Series

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

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

= +

1, 2, 4, …, 64 Finite Geometric( )n

1n

a r 1S

r 1−

=−

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

No Sum

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

Infinite Geometric-1 < r < 1

1aS1 r

=−

Find the sum, if possible: 1 1 11 ...2 4 8

+ + + +

1 112 4r

11 22

= = = 1 r 1 Yes→ − ≤ ≤ →

1a 1S 211 r 12

= = =− −

What? If possible? What are they talking about?

If r is between -1 and 1!

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

8 16 2r 2 282 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: 510 5 ...2

+ + +

55 12r

10 5 2= = = 1 r 1 Yes→ − ≤ ≤ →

1a 10S 2011 r 12

= = =− −

Converting repeating decimals to fractions

Write the repeating decimal 0.808080… as a fraction.

Write the repeating decimal 0.153153… as a fraction.

The Bouncing Ball Problem – Version A

A ball is dropped from a height of 50 feet. It rebounds 4/5 ofit’s height, and continues this pattern until it stops. How fardoes the ball travel?

50

40

32

32/5

40

32

32/5

40S 455041

01

554

= =−

+−

The Bouncing Ball Problem – Version B

A ball is thrown 100 feet into the air. It rebounds 3/4 ofit’s height, and continues this pattern until it stops. How fardoes the ball travel?

100

75

225/4

100

75

225/4

10S 80100

4 431

0

10

3= =

−+

−

Sigma Notation Section 12-5

B

nn A

a=∑

UPPER BOUND(NUMBER)

LOWER BOUND(NUMBER)

SIGMA(SUM OF TERMS) NTH TERM

(SEQUENCE)

( )j

4

1j 2

=

+∑ ( )21= + ( )2 2+ + ( )3 2+ + ( )24+ + 18=

( )7

4a2a

=∑ ( )( )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=

0

n

b

365=

∞ ⎛ ⎞ =⎜ ⎟⎝ ⎠

∑036

5⎛ ⎞⎜ ⎟⎝ ⎠

1365

⎛ ⎞+ ⎜ ⎟⎝ ⎠

2365

⎛ ⎞+ ⎜ ⎟⎝ ⎠

...+

1aS1 r

=−

6 15315

= =−

( )2

x

3

72x 1

=

+∑ ( )( ) ( )( ) ( )( ) ( )( )2 1 2 8 1 2 9 1 ...7 2 123= + + + + + + + +

( ) ( )n 1 n2n 1S a a 15

23

27 47− +

= + = + 527=

( )1

b

9

44b 3

=

+∑ ( )( ) ( )( ) ( )( ) ( )( )4 3 4 5 3 4 6 3 ...4 4 319= + + + + + + + +

( ) ( )n 1 n1n 1S a a 19

29

24 79− +

= + = + 784=

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

Arithmetic, d= 3

( )n 1a a n 1 d= + −

( )na 3 n 1 3= + −

na 3n=4

1n3n

=∑

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

Geometric, r = ½n 1

n 1a a r −=n 1

n1a 162

−⎛ ⎞= ⎜ ⎟⎝ ⎠

n 1

n

5

1

1162

−

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

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

Not Arithmetic, Not Geometric

n 1na 20 2 −= −

n 1

n

5

120 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 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 −∞

=∑