arithmetic sequences and series ncs mathematics dvd series resources... · 2020. 4. 8. · lesson 1...
TRANSCRIPT
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NCS Mathematics DVD Series
Arithmetic Sequences
and Series
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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
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NCS Mathematics DVD Series
Lesson 1
Number Sequences and
Generating Rules
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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
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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
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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
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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� � �
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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;
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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�
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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
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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� �
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NCS Mathematics DVD Series
Lesson 2
Arithmetic Sequence
and the General Term
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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
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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 ; � � � �
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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 � � � �
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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
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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
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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.
PAUSE DVD
• Do Tutorial 2
• Then View solutions
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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? � � � �
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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� �
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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
� � �� � � �
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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
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NCS Mathematics DVD Series
Lesson 3
Arithmetic Series and
Sigma Notation
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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:
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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
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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
? � � � � �
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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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• Do Tutorial 3
• Then View Solutions
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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 � � � � � � � � � � ¦
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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!!)
� � � �
�
¦
¦
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NCS Mathematics DVD Series
Lesson 4
The Sum of an Arithmetic
Series
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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
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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.
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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
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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
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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
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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
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108 3 3 1020
3 111 1020 0
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2 37 340 017 20 0
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> @ � �20(b) Sum of the terms 4 99 10 103 1030.2
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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.
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nT n � �
So 4 5 5 99 5 100 20.n n n� � � � Hence there are 20 terms in the series.
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(b) Total earnings after 10 years:10 2 90000 9 45002
5 220500 1 102 500.
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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 �
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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!