NCS Mathematics DVD Series

Arithmetic Sequences

and Series

Outcomes for this DVD

In this DVD you will:

• Investigate number sequences and generating rules. LESSON 1 • Identify arithmetic sequences and construct the formula for the general term. LESSON 2 • Discuss the arithmetic series and sigma notation. LESSON 3 • Construct and apply the sum formula of an arithmetic series. LESSON 4

NCS Mathematics DVD Series

Lesson 1

Number Sequences and

Generating Rules

Number Sequences

DEFINITION: A number sequence is a list of numbers given in a particular order.

Numbers in a sequence may be generated randomly or by a particular rule

EXAMPLE 1: 1; 2; 4; 8; …

EXAMPLE 2: 18; 13; 8; 3; -2; …

EXAMPLE 3: 1; 4; 9; 16; …

EXAMPLE 4: 1; 8; 27; 64; …

Rule: Multiply by 2

Rule: Subtract 5 or add -5

Rule: Square a number

Rule: Cube of a number

Tutorial 1: Identifying Rules for Number Sequences

• Identify the rule/pattern in the following sequences • Write down the next three terms of each sequence

1) 5; 15; 45; … 2) 7; 4; 1; -2; … 3) 1; 1; 2; 3; 5; … 4) 2; 5; 10; 17; … 5) 27; 9; 3; 1; … 6) 0; 7; 26; 63; …

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Tutorial 1 Problem 1: Suggested Solution

1) 5; 15; 45;

Generating Rule: Multiply by 3

Next three terms:45 3 135;u 135 3 405;u 405 3 1 215u

Tutorial 1 Problem 2: Suggested Solution

2) 7; 4; 1; 2; �

Generating Rule: Add 3 or Subtract 3�Next three terms:

� �2 3 5;� � � �

� �5 3 8;� � � �

� �8 3 11� � �

Tutorial 1 Problem 3: Suggested Solution

3) 1; 1; 2; 3; 5;

Generating Rule:Add the two preceeding numbers

Next three terms:3 5 8;� 8 5 13;� 13 8 21�

1; 1; 2; 3; 5; 8; 1; 1; 2; 3; 5; 8; 13; 1; 1; 2; 3; 5; 8; 13; 21;

Tutorial 1 Problem 4: Suggested Solution

4) 2; 5; 10; 17; Generating Rule:Square Natural Numbers and Add 1Next three terms:

25 1 26;� 26 1 37;� 27 1 50�

Tutorial 1 Problem 5: Suggested Solution

5) 27; 9; 3; 1; Generating Rule:

1Divide by 3 or Multiply by 3

Next three terms:11 3 ;3

y 1 1 1 ;3 3 9u

1 1 19 3 27u

Tutorial 1 Problem 6: Suggested Solution

6) 0; 7; 26; 63; Generating Rule:Cube the Natural Numbers minus 1Next three terms:

35 1 125 1 124;� � 36 1 216 1 215;� � 37 1 343 1 342� �

NCS Mathematics DVD Series

Lesson 2

Arithmetic Sequence

and the General Term

Common Difference implies an Arithmetic Sequence

The fixed number that we add is called the common difference and is generally denoted by d.

Consider the following sequences:

2; 7; 12; 17; …

8; 2; -4; -10; …

6; 15; 24; 33; …

6� 6� 6�

5 5 5

9 9 9 Common Difference Arithmetic Sequence

 

n n 1Td T �? �

2 1 3 2 4 3 T T T T T Td � � � Common Difference

General Term of an Arithmetic Sequence (AS)

Then the sequence is:

st

thn

Let be the 1 term, the constant differenceand the n term of an Arithmetic SequT

daence

1 d0aT � u

2T 1a d � u

3T 2a d � u

10 a T 9

d � u nT (a n 1)d � �

thHence the general n termof an AS is given by:

Explicit Form

a a a a; 2 3d; d; d; 4 d a ; � � � �

Problem 1 on Arithmetic Sequences

Example 5

Hence the 19th term is -39 .

th

For a sequence 15; 12; 9; ... , determine (a) the 11 term and(b) which term is 39�

Solution

Here 15 and a 3d �

So15 ( 1)( 3) 3915 3 3 393 57

19

� � � �� � � �� �

nn

nn

� �b Let the term be 39. Then ( 1) 39.

thnna d

�

� � �

� � 11a 10 15 (10)( 3) 15.daT � � � �

Problem 2 on Arithmetic Sequences

Example 6

.

Solution

The 4th term of an AS is 14 and the 16th term is 50. Calculate the first term and the common difference.

4The 4 term 3 14. (1) � th aT d

16The 16 term 15 50. (2) � th a dT

Subtracting equation (1) from equation (2), we get 12d d36 3 �

So 3(3) 14 5� � a a

Recursive Formula for an Arithmetic Sequence

Each new term in an arithmetic sequence comes from adding the common difference d to the previous term. 1 a T

2 d a T �

3 2d dT 2 Ta � �

10 9

T 9 T

da d � �

n

n 1

d T (n 1)a

dT �

� � �

Consider the terms of an AS:

fHence a recursive ormulafor an AS is given by:

1

1 ; 1n n

Tn

aT dT �

� t

1 1 ; 1n na T dT T n� � !Not unique!

and

Tutorial 2: Questions on Arithmetic Sequences

2; 4 ; 6 4.

25; 14; 313

52.

(1) are three consecutive terms of an AS.Determine the value of

(2) If ; ... forms an AS, determine (a) the term, (b) a formula for the general term, (c) which term is

� �

�

th

x x xx

1 1

4 20.

410

1 3 1

th

k kT T T k�

� �

� t

(3) The third term of an AS is and the seventh term is Determine

(a) the first terms, (b) the term(4) A sequence is defined by and for .

Determine the first four terms of the sequence.

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Tutorial 2 Problem 1: Suggested Solution

3 2 2 4 6x x x? � � � Three consecutive terms are 8; 24 and 40?

Given 2; 4 ; 6 4 as three consecutiveterms for an AS. Determine value of .

x x xx

� �

2 1 3 2For an AS, T T T T� �

(4 ) ( 2) (6 4) (4 )x x x x? � � � �

Tutorial 2 Problem 2: Suggested Solution

2(b) General term: ( 1) 25 ( 1)( 11) 36 11

nT a n dnn

� � � � � �

Then 25 and 11.2(a) 13 term 12 25 12( 11) 107th

a da d

�

? � � � �

Hence the 8 term is 52.th �

Given sequence 25;14; 3; as an AS:

2(c) Suppose that 52.nT � Then 36 11 52n� �11 88 8.n n� �

Tutorial 2 Problem 3: Suggested Solution

(a) First four terms are: 4; 0; 4; 8.(b) 10 term 9 4 9( 4) 32.th a d

� �

� � � �

(3) 3 term 4 2 4 ... (1) 7 term 20 6 20 ... (2)

rd

th

a da d

� � � �

� � � �Subtracting equation (1) from equation (2), we get

4 16 4So 2( 4) 4 4.

d da a

� � �� � � �

Tutorial 2 Problem 4: Suggested Solution

1

2 1

3 2

4 3

(4) 13 1 3 43 4 3 73 7 3 10

TT TT TT T

� � � � � �

1 11 3 1k kT T T k� � tGiven and for

NCS Mathematics DVD Series

Lesson 3

Arithmetic Series and

Sigma Notation

Arithmetic Series

1 2 n

1 2 n

T ;T ;...T

T T ... T

Let denote the terms of an Arithmetic Sequence.Then the sum of the terms of the sequence:

Arithmetic Seriesis called an � � �

Definition

Example 7

If we add a finite number of terms, the series is called finite.

nn SIf the first terms of an AS are added, the sum is denoted by .

4 5 9 13 17 44 � � � S7 5 9 13 17 21 25 29 119S � � � � � �

5; 9; 13; 17; 21;For the AS:

Sigma Notation and Series Expansion

1 2 nn

1 2 n kk 1

k

T T ... T

T T ... T T

T

A series can be written in as follows:

where is the general term of the c

Sigma N

orrespo

ota

ndi

t

n

io

g

n

sequence.

� � �

� � � ¦

Example 8

> @ > @ > @ > @ > @5

1(4 1) 4( ) 1 4( ) 1 4( ) 1 41 2 3 4 5( ) 1 4( ) 1

3 7 11 15 19 55r

r

� � � � � � � � � �

� � � �

¦

n

k kk 1

T T , k 1 n.Read: the sum of the terms for to

� ¦

62 2 2 2 2

33 4 5 6 9 16 25 36 86

� � � � � � ¦kk

Expanding a series from Sigma Notation

Choice of

variable arbitrary

Writing a Series in Sigma Notation To write a given series in sigma notation is more complicated.For this k

we need the formula for the general term T .

Example 9

102 2 2 2 2 2

1(a) Clearly for 1 to 10. Hence 1 2 3 ... 10

� � � � ¦kk

T k k k

5 5

1 1Hence 5 11 17 23 29 (6 1)k

k kT k

� � � � �¦ ¦

Write the series in Sigma Notation2 2 2 2(a) 1 2 3 ... 10� � � � (b) 5 11 17 23 29� � � �

Solution

� � � �(b) This is an AS with five terms and 5, 6. 1 5 1 6 6 1kT a k d k

a dk

? � � � � �

Tutorial 3: Arithmetic Series and Sigma Notation

4 5

1 05 20

1

2 1

( 1) (b) 2

(c) ( 1) (d) 5

�

�

�

¦ ¦

¦ ¦

k

k k

r

r k

k k

(1) Expand and evaluate the following:

(a)

3 3 3 3

:1 2 3 ... 7

( 3 4 5 ... 188 13 18 23 28

� � � �� � � �� � � �

(2) Write the following in sigma notation(a) b)

(c)

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Tutorial 3 Problem 1: Suggested Solutions

� � � � � � � �4

11(a) ( 1) 1 2 2 3 3 4 4 5 40

kk k

� � � � ¦

� �20 20

0

1 1 (d) 5(1) 5( ) 5 5 5 ... 5 20 5 100

k kk

� � � � ¦ ¦

� � � � � � � �� � � �

51

23 4 5 6

(c) ( 1)

1 1 1 1

1 1 1 1 0

r

r

�

�

� � � � � � �

� � � � �

¦

50 1 2 3 4 5

0 (b) 2 2 2 2 2 2 2 1 2 4 8 16 32 63k

k � � � � � � � � � � ¦

Tutorial 3 Problem 2: Suggested Solution

k5

k 1

T 8 (k 1)5 5k 3;

k 1 5. 8 13 18 23 28 (5k 3).

(c) This is an Arithmetic series with

where to Hence

� � �

� � � � �¦

3k

73 3 3 3 3

k 1

(2) T k , k 1 7.

1 2 3 ... 7 k .

(a) By inspection for to

Hence

� � � � ¦k

18

k 3

16

k 1

T k, k 3 18.

3 4 5 ... 18 k.

(k 2)

(b) By inspection, for to

Hence

Alternative Solution: (verify!!)

� � � �

�

¦

¦

NCS Mathematics DVD Series

Lesson 4

The Sum of an Arithmetic

Series

Summing the terms of an Arithmetic Series

> @ > @ � � � �2 ... 2 1 ... (1)ª º ª º � � � � � � � � � � �¬ ¼ ¬ ¼nS a a d a d a n d a n d

� � � � > @ > @1 2 ... 2 ... (2)ª º ª º � � � � � � � � � � �¬ ¼ ¬ ¼nS a n d a n d a d a d a

1 d.T a Consider an Arithmetic Series with and constant differen e c

Then

Reversing the order of terms in (1), we get:

� � � � � � � �2 2 1 2 1 ... 2 1 2 1nS a n d a n d a n d a n dª º ª º ª º ª º � � � � � � � � � � � �¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

Summing the corresponding terms in (1) and (2) gives us:

> @2 2 ( 1)� � �nS n a n d

� � Hence 2 12ª º � �¬ ¼n

nS a n d� �

> @

Last term ( 1)

Hence 12

2

� �

ª º � � �¬ ¼

�

n

n

L nnS a a n d

nS

a

a

d

L

Note

Finding the sum of an Arithmetic Series Example 10

5; 8; 11; 14; ...,

440.

For the Arithmetic sequence determine(a) the sum of the first 30 terms,(b) the number of terms that must be added for a sum total of

� � � � � �3030(a) 2 5 29 3 15 97 1455 2ª º � ¬ ¼SSolution

� � � �

� �� �

2

(b) Suppose terms are added. Then 440.

2 5 1 3 44023 7 880 0

16 3 55 05516 or 3

nn Sn n

n nn n

n n

� � � ª º¬ ¼

� � �

� � �

�� 16 Hence terms must be added.

Finding the Sum of an Arithmetic Series Example 11

7 11 15 ... 151For the Arithmetic series determine(a) the number of terms,(b) the sum all the terms.

� � � �

� �(a) Suppose the last term is the term. Then

7 1 4 151 7 4 4 151 4 148 37.Hence there are 37 terms in the series. � � � � � � �

th

n

nT n n n n

Solution

> @3737(b) The sum of all the terms is 7 151 2923. 2

� S

Tutorial 4: Sums of Arithmetic Sequences

(4) A job was advertised at a starting salary of R90 000 pa with an annual increase of R4 500. Determine: (a) the employee's salary in the sixth year, (b) the total earnings after 10 years.

(1) Calculate the sum of: (a) the first 18 terms of the sequence: 7; 13; 19; ... (b) the first 7 terms of the sequence: 24; 21; 18; ...(2) Determine how many terms of the sequence 54; 51; 48; ... must be added to make a sum total of 510.(3) For a series 4 9 14 ... 99, determine: (a) the number of terms, (b) the sum of the terms.

� � � �PAUSE DVD • Do Tutorial 4 • Then View Solutions

Tutorial 4 Problem 1: Suggested Solutions

� � � �1818(1) (a) 2 7 17 6 9(116) 10442

S � ª º¬ ¼

(1) Calculate the sum of: (a) the first 18 terms of the sequence: 7; 13; 19; ... (b) the first 7 terms of the sequence: 24; 21; 18; ...

� � � � � �77 7(1) (b) 2 24 6 3 30 1052 2

S � � ª º¬ ¼

Tutorial 4 Problem 2: Suggested Solutions

Hence summing either 17 or 20 terms will result in a total of 510.

(2) Determine how many terms of the sequence 54; 51; 48; ... must be added to make a sum total of 510.

(2) If terms are added then 510.nn S

� � � �� �So 2 54 1 3 5102n n� � � ª º¬ ¼

> @2

108 3 3 1020

3 111 1020 0

n n

n n

� � �

� � �

� �� �

2 37 340 017 20 0

n nn n

� � �

� � � 17 or 20n n�

> @ � �20(b) Sum of the terms 4 99 10 103 1030.2

�

Tutorial 4 Problem 3: Suggested Solutions

(3) For a series 4 9 14 ... 99, determine: (a) the number of terms, (b) the sum of the terms.

� � � �

� �(3) (a) Suppose there are terms. Then 4 1 5 99.n

nT n � �

So 4 5 5 99 5 100 20.n n n� � � � Hence there are 20 terms in the series.

� � � �

� �10

(b) Total earnings after 10 years:10 2 90000 9 45002

5 220500 1 102 500.

S

R

�ª º¬ ¼

Tutorial 4 Problem 4: Suggested Solutions

(4) A job was advertised at a starting salary of R90 000 pa with an annual increase of R4 500. Determine: (a) the employee's salary in the sixth year, (b) the total earnings after 10 years.

6(4) (a) Salary in the sixth year 90000 5(4500) 112 500.T R �

End of the DVD on Arithmetic Sequences

REMEMBER! • Consult text-books for additional examples.

• Attempt as many as possible other similar examples on your own.

• Compare your methods with those that were discussed in the DVD.

• Repeat this procedure until you are confident.

• Do not forget:

Practice makes perfect!