# Arithmetic Sequences and Series NCS Mathematics DVD Series resources... · 2020. 4. 8. · LESSON 1 • Identify arithmetic sequences and construct the formula for the general term

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• 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; …

PAUSE DVD

• Do Tutorial 1

• Then View Solutions

• Tutorial 1 Problem 1: Suggested Solution

1) 5; 15; 45;

Generating Rule: Multiply by 3

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

• Tutorial 1 Problem 2: Suggested Solution

2) 7; 4; 1; 2;

Generating Rule: Add 3 or Subtract 3Next 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

1 1 1 ;3 3 9

1 1 19 3 27

• 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 TdCommon 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

2T 1a d

3T 2a d

10 a T 9

d

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 5a 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

1 1 ; 1n na T dT T nNot 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

(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.

PAUSE DVD

• Do Tutorial 2

• Then View solutions

• Tutorial 2 Problem 1: Suggested Solution

3 2 2 4 6x x xThree 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 52n11 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 get4 16 4

So 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 kGiven 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 ... 10k

kT 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)

PAUSE DVD

• Do Tutorial 3

• Then View Solutions

• Tutorial 3 Problem 1: Suggested Solutions

4

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

kk k

20 200

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

k18

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)nS a a d a d a n d a n d

1 2 ... 2 ... (2)nS a n d a n d a d a d a

1 d.T aConsider 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 d

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

2 2 ( 1)nS n a n d

Hence 2 12nnS a n d

Last term ( 1)

Hence 12

2

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 2

SSolution

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 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

(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

• 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

2

108 3 3 1020

3 111 1020 0

n n

n n2 37 340 0

17 20 0n 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 nHence 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

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:

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