An introduction…………

1, 4, 7,10,13

9,1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

− −

π π + π +

Arithmetic Sequences

ADD

To get next term

2, 4, 8,16, 32

9, 3,1, 1/3

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

, 2.5 , 6.25

− −

π π π

Geometric Sequences

MULTIPLY

To get next term

Arithmetic Series

Sum of Terms

35

12

27.2

3 9

−

π +

Geometric Series

Sum of Terms

62

20/3

85/ 64

9.75π

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 = 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

n

S a a

2

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

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

n

S a a

2

= +

( )63

63

3 3S

2

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

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=

13

x

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

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 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

n

S a a

2

= +

( )y 5 31x= + −

( )

x

40 y4

2

5= +

( )( )1

2

x

440 5 5 x 3= + + −

( )x 7 x

440

2

3

=

+

( )880 x 7 3x= +

2

0 3x 7x 880= + −

X = 16

Graph on positive window

The sum of the first n terms of an infinite sequence

is called the nth partial sum.

1( )

2n n

nS a a= +

Example 6. Find the 150th

partial sum of the arithmetic sequence, 5,

16, 27, 38, 49, …

1 5 11 5 11 6a d c= = → = − = −

11 6na n= − ( )150 11 150 6 1644a→ = − =

( ) ( )150

150

5 1644 75 1649 123,675

2

S = + = =

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?

1 20 1 19d c= = − =

( ) ( )1 201 20 19 1 39na a n d a= + − → = + =

( ) ( )20

20

20 39 10 59 590

2

S = + = =

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.

1 10,000 7500 10,000 7500 2500a d c= = = − =

( ) ( )1 201 10,000 19 7500 152,500na a n d a= + − → = + =

( ) ( )20

20

10,000 152,500 10 162,500 1,625,000

2

S = + = =

So the total sales for the first 2o years is $1,625,000

1, 4, 7,10,13

9,1, 7, 15

6.2, 6.6, 7, 7.4

, 3, 6

− −

π π + π +

Arithmetic Sequences

ADD

To get next term

2, 4, 8,16, 32

9, 3,1, 1/3

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

, 2.5 , 6.25

− −

π π π

Geometric Sequences

MULTIPLY

To get next term

Arithmetic Series

Sum of Terms

35

12

27.2

3 9

−

π +

Geometric Series

Sum of Terms

62

20/3

85/ 64

9.75π

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

n

1

n

nth term of geometric sequence

sum of n terms of geometric sequ

a a r

a r 1

S

r 1

ence

−

→ =

− =

−

→

1 9

1 2

If 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 1

n 1a a r −

=

9 1

1 2

x

2 3

−

=

8

8

2

x

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 1

n 1a a r −

=

( ) ( )

14

54 2 x

−

−=

3

27 x− =

3 x− =

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

forms an geometric sequence

2 4 1

2

Find a a if a 3 and r

3

− = − =

-3, ____, ____, ____

2

Since r …

3

=

4 8

3, 2, ,

3 9

− −

− −

2 4

8 10

a 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 2

r 2

22

= = =

n 1

n 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

2−

n 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

x

n 1

n 1a a r −

=

3 11

4

4

r −

= 2

r

1

4

4

→ = 2

16 r→ = 4 r→ ± =

1

,1, 4

4

1

, 1, 4

4

−

7

1 1 1

Find 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

11

184r

1 1 2

2 4

= = =

( )n

1

n

a r 1

S

r 1

− =

−

7

1 1

2 2

x

1

2

1

1

− =

−

7

1 1

2 2

1

2

1

− =

−

63

64

=

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

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

n

S a a

2

= +

1, 2, 4, …, 64 Finite Geometric

( )n

1

n

a r 1

S

r 1

−

=

−

1, 2, 4, 8, … Infinite Geometric

r > 1

r < -1

No Sum

1 1 1

3,1, , , …

3 9 27

Infinite Geometric

-1 < r < 1

1a

S

1 r

=

−

Find the sum, if possible:

1 1 1

1 …

2 4 8

+ + + +

1 1

12 4r

11 2

2

= = = 1 r 1 Yes→ − ≤ ≤ →

1a 1

S 2

11 r

1

2

= = =

−

−

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

8 16 2

r 2 2

82 2

= = = 1 r 1 No→ − ≤ ≤ →

NO SUM

Find the sum, if possible:

2 4 8

…

7 7 7

+ + +

4 8

7 7r 2

2 4

7 7

= = = 1 r 1 No→ − ≤ ≤ →

NO SUM

Find the sum, if possible:

5

10 5 …

2

+ + +

5

5 12r

10 5 2

= = = 1 r 1 Yes→ − ≤ ≤ →

1a 10

S 20

11 r

1

2

= = =

−

−

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

40

S 45

50

4

1

0

1

55

4

= =

−

+

−

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

10

S 80

100

4 4

3

1

0

1

0

3

= =

−

+

−

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

1n

3n

=

∑

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

Geometric, r = ½

n 1

n 1a a r −

=

n 1

n

1

a 16

2

−

=

n 1

n

5

1

1

16

2

−

=

∑

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

Not Arithmetic, Not Geometric

n 1

na 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 = 3

Denominator 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

−

∞

=

∑

48

Do you find this slides were useful?

One second of your life , can bring a smile in a girl life

If Yes ,Join Dreams School “Campaign for Female Education”

Help us in bringing a change in a girl life, because “When

someone takes away your pens you realize quite how

important education is”.

Just Click on any advertisement on the page, your one

click can make her smile.

Eliminate Inequality “Not Women”

One second of your life , can bring a smile in her life!!

Do you find these slides were useful?

If Yes ,Join Dreams School “Campaign

for Female Education”

Help us in bringing a change in a girl life,

because “When someone takes away your

pens you realize quite how important

education is”.

Just Click on any advertisement on the page,

your one click can make her smile.

We our doing our part & u ?

Eliminate Inequality “Not Women”