Click here to load reader

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

  • View
    0

  • Download
    0

Embed Size (px)

Text of Arithmetic Sequences and Series NCS Mathematics DVD Series resources... · 2020. 4. 8. · LESSON 1...

  • 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:

    Practice makes perfect!