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KS4 Mathematics

KS4 Mathematics. A7 Sequences. A7 Sequences. Contents. A. A7.2 Linear sequences. A. A7.1 Generating sequences from rules. A7.3 Quadratic sequences. A. A7.4 Geometric sequences. A. A7.5 Other types of sequence. A. Sequences. 1 st term. 6 th term.

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KS4 Mathematics

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  1. KS4 Mathematics A7 Sequences

  2. A7 Sequences Contents • A A7.2 Linear sequences • A A7.1 Generating sequences from rules A7.3 Quadratic sequences • A A7.4 Geometric sequences • A A7.5 Other types of sequence • A

  3. Sequences 1st term 6th term 2, 5, 8, 11, 14, 17, 20, 23, . . . In mathematics, a sequence is a succession of numbers that follow a given rule. Each number in a sequence is called a term. For example, If terms are next to each other they are referred to as consecutive terms.

  4. Predicting terms in a sequence For example, What are the next two terms in the following sequence, 102, 95, 88, 81, 74 . . . ? –7 –7 –7 –7 –7 –7 102 95 88 81 74 Sometimes, we can predict how a sequence will continue by looking for patterns. Look at the difference between each consecutive term. 67 60 We can predict that this sequence continues by subtracting 7 each time. We can use this to find the next two terms.

  5. Predicting terms in a sequence +2 +4 +6 +8 +10 +12 2 4 8 ×2 ×2 ×2 ×2 ×2 ×2 2 4 8 If we are not given the rule for a sequence, or if it is not generated from a practical context, we cannot be certain how it will continue. For example, a sequence starts with the numbers 2, 4, 8, . . . How could this sequence continue? 14 22 32 44 16 32 64 128

  6. Defining sequences The first is to use a term-to-term rule. The second is to use a position-to-term rule. There are two ways to define a sequence. To define a sequence using a term-to-term rule we need to know the first term in the sequence and what must be done to each term to give the value of the next term. To define a sequence using a position-to-term rule we use a formula for the nth term of the sequence. Term-to-term rules are usually easier to find for a given sequence. Position-to-term rules are harder to find for a given sequence but are more useful for finding any term in a sequence.

  7. Sequences from term-to-term rules Write the first five terms of each sequence given the first term and the term-to-term rule. 1st term Term-to-term rule First five terms –7 Add 3 –7, –4, –1, 2, 5 100 Subtract 8 100, 92, 84, 76, 68 3 Double 3, 6, 12, 24, 48 Multiply by 10 0.04, 0.4, 4, 40, 400 0.04 1.3 Subtract 0.6 1.3, 0.7, 0.1, –0.5, –1.1 111 Divide by 5 111, 22.2, 4.44, 0.888, 0.1776

  8. Sequences from position-to-term rules Position 1st 2nd 3rd 4th 5th … nth Term … n2 When a sequence is defined by a position-to-term rule it can sometimes help to put the terms in a table. For example, The nth term in a sequence is n2, where n is the term’s position in the sequence. 1 4 9 16 25 Each term can be found by squaring its position in the sequence. What is the 20th term in this sequence? 202 = 400

  9. Writing sequences from position-to-term rules The nth term, or the general term, of a sequence is often called un. The 1st term is then called u1, the 2nd term u2, the 3rd term u3, the 4th term u4, the 5th term u5 and so on. Any term in the sequence can then be found by substituting its position number into the formula for un.

  10. Writing sequences from position-to-term rules For example, suppose the nth term of a sequence is 4n – 5. We can write this rule as: un = 4n – 5 Find the first five terms in the sequence. 4 ×1 – 5 = –1 u1 = 4 ×2 – 5 = 3 u2 = 4 ×3 – 5 = 7 u3 = 4 ×4 – 5 = 11 u4 = 4 ×5 – 5 = 15 u5 = The first five terms in the sequence are: –1, 3, 7, 11 and 15.

  11. Writing sequences from position-to-term rules The nth term of a sequence is 2n2 + 3. We can write this rule as: un = 2n2 + 3 Find the first 4 terms in the sequence. 2 ×12 + 3 = 5 u1 = 2 ×22 + 3 = 11 u2 = 2 ×32 + 3 = 21 u3 = 2 ×42 + 3 = 35 u4 = The first 4 terms in the sequence are: 5, 11, 21, and 35. This sequence is a quadratic sequence.

  12. Writing sequences from position-to-term rules

  13. Matching sequences to rules

  14. A7 Sequences Contents A7.1 Generating sequences from rules • A • A A7.2 Linear sequences A7.3 Quadratic sequences • A A7.4 Geometric sequences • A A7.5 Other types of sequence • A

  15. Sequences that increase in equal steps +6 +6 +6 +6 +6 +6 +6 To work out a rule for a sequence it is often helpful to find the difference between consecutive terms. For example, look at the difference between each term in this sequence: 2, 8, 14, 20, 26, 32, 38, 44, . . . This sequence starts with 2 and increases by adding 6 each time. Every term in this sequence is 4 less than a multiple of 6.

  16. Sequences that decrease in equal steps –3 –3 –3 –3 –3 –3 –3 Can you work out the next three terms in this sequence? 10, 7, 4, 1, –2, –5, –8, –11, . . . How did you work these out? This sequence starts with 10 and decreases by subtracting 3 from each term to give the next term. When the difference between each term in a sequence is always the same we have a linear or arithmetic sequence.

  17. Generating linear sequences When we describe linear sequences we call the constant difference between consecutive terms the common difference, d. We call the first term in a linear sequence a. If we are given the values of a and d in a linear sequence then we can use them to generate the sequence. For example, if a linear sequence has a = 5 and d = –3, we have the sequence: 5, 2, –1, –4, –7, –10, . . . Linear sequences can also be generated given the rule for the nth term in the sequence.

  18. The nth term of a linear sequence × 3 × 3 × 3 × 3 + 1 × 3 + 1 + 1 + 1 + 1 × 3 + 1 Suppose we are given a linear sequence and asked to find the nth term, un, of the sequence. For example, Find the nth term of the sequence 4, 7, 10, 13, 16, … This sequence continues by adding 3 each time and so the common difference d is 3. We compare the terms in the sequence to the multiples of 3. 2 3 4 5 n Position 1 … Multiples of 3 3 6 9 12 15 3n Term 4 7 10 13 16 … 3n + 1 un = 3n + 1.

  19. The nth term of a linear sequence × –2 × –2 × –2 × –2 + 7 × –2 + 7 + 7 + 7 + 7 × –2 + 7 Find the nth term of the sequence 5, 3, 1, –1, –3, … This sequence continues by subtracting 2 each time and so the common difference d is –2. We compare the terms in the sequence to the multiples of –2. 2 3 4 5 n Position 1 … Multiples of –2 –2 –4 –6 –8 –10 –2n Term 5 3 1 –1 –3 … 7 – 2n un = 7 – 2n.

  20. The nth term of a linear sequence If a is the first term of a linear sequence and d is the common difference, we can find the general form of the nth term as follows: u1 = a, u2 = a + d, u3 = a + 2d, u4 = a + 3d, un= a + (n – 1)d, Multiplying out the bracket we have un= a + dn – d. The nth term of a linear sequence with first term a and common difference d is: (a – d) is the value of the 0th term. un= dn + (a – d)

  21. Plotting terms Value of term (un) Position number (n) We can also find the nth term of a linear sequence by plotting the value of each given term in the sequence against its position number. For example, The sequence 3, 5, 7, 9, 11 … can be shown graphically: The points lie on a straight line. The equation of the line can be written in the form y = mx + c. The value of the gradient m corresponds to the difference between the terms. The value of the y-intercept c corresponds to the 0th term, to give un=2n + 1

  22. Finding the nth term of a linear sequence

  23. Tiling patterns Pattern 1 Pattern 2 1 tile Pattern 3 5 tiles Pattern 4 9 tiles 13 tiles The following patterns are made from tiles: How many tiles will there be in the next pattern? How many tiles will there be in the 20th pattern?

  24. Tiling patterns +4 +4 +4 +4 To work out the number of tiles in the next pattern we can look at the difference between consecutive terms in the sequence: 1, 5, 9, 13, 17 The difference between each term is always equal to 4 and so we can use this to find the next term. Using this pattern, we can predict that there will be 17 tiles in the next pattern.

  25. Tiling patterns +4 +4 +4 +4 To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence. Look at the pattern made by the differences: 1, 5, 9, 13, 17 This is a linear sequence and so the nth term will be of the form un = dn + c, where d is the constant difference between the terms. For this sequence the constant difference d = 4. How can we find the value of c?

  26. Tiling patterns 2 3 4 5 n Position 1 … Term 1 5 9 13 17 … – 3 – 3 × 4 × 4 × 4 × 4 × 4 – 3 × 4 – 3 – 3 – 3 We can find the rule for the nth term by comparing the sequence to multiples of 4. Multiples of 4 4 8 12 16 20 4n 4n – 3 un = 4n – 3. We can use this formula to predict the number of tiles in the 20th pattern: u20 = 4 × 20 – 3 = 77

  27. A7 Sequences Contents A7.1 Generating sequences from rules • A A7.2 Linear sequences • A A7.3 Quadratic sequences • A A7.4 Geometric sequences • A A7.5 Other types of sequence • A

  28. Sequences that increase in increasing steps +1 +2 +3 +4 +5 +6 +7 Some sequences increase or decrease in unequal steps. For example, look at the differences between terms in this sequence: 4, 5, 7, 10, 14, 19, 25, 32, . . . This sequence starts with 4 and increases by adding consecutive whole numbers to each term. The differences between the terms form a linear sequence.

  29. Sequences that decrease in decreasing steps –0.1 –0.2 –0.3 –0.4 –0.5 –0.6 –0.7 Can you work out the next three terms in this sequence? 7, 6.9, 6.7, 6.4, 6, 5.5, 4.9, 4.2, . . . How did you work these out? This sequence starts with 7 and decreases by subtracting 0.1, 0.2, 0.3, 0.4, 0.5, … With sequences of this type it is often helpful to find a second row of differences.

  30. Using a second row of differences +20 +17 +11 +8 +5 +2 +3 +3 +3 +3 +3 +14 +3 Can you work out the next three terms in this sequence? 1, 3, 8, 16, 27, 41, 58, 78 First row of differences Second row of differences Look at the differences between terms. A sequence is formed by the first row of differences so we look at the second row of differences. This shows that the differences increase by 3 each time.

  31. Quadratic sequences +3 +5 +7 +9 +2 +2 +2 When the second row of differences produces a constant number, the sequence is called a quadratic sequence. This is because the rule for the nth term of the sequence is a quadratic expression of the form un = an2 + bn + c where a, b and c are constants and a≠ 0. The simplest quadratic sequences is the sequence of square numbers. 1, 4, 9, 16, 25 The constant second difference is 2 and the nth term isn2.

  32. Investigating quadratic sequences

  33. The nth term of a quadratic sequence When un = an2 + bn + c the value of a can be found by halving the value of the second difference, which is always equal to 2a. We can prove this as follows, un= an2 + bn + c u1 = a × 12 + b × 1 + c = a + b + c u2 = a × 22 + b × 2 + c = 4a + 2b + c u3 = a × 32 + b × 3 + c = 9a + 3b + c u4 = a × 42 + b × 4 + c = 16a + 4b + c u5 = a × 52 + b × 5 + c = 25a + 5b + c This gives us the first five terms of a general quadratic sequence.

  34. The nth term of a quadratic sequence Let’s find the first and second differences for these general terms. a + b + c 4a + 2b + c 9a + 3b + c 16a + 4b + c 25a + 5b + c 3a + b 5a + b 7a + b 9a + b 2a 2a 2a This shows that the second difference is always 2a when un = an2 + bn + c.

  35. The nth term of a quadratic sequence When un = an2 + bn + c the value of c can be found by calculating the value of the 0th term, u0. We can prove this as follows, un= an2 + bn + c u0 = a × 02 + b × 0 + c u0 = c Of course, the 0th term is not really part of the sequence. We can easily work it out though by looking at the sequence formed by the first row of differences and counting back from the first term.

  36. The nth term of a quadratic sequence When un = an2 + bn + c the value of b can be found by subtracting (a + c) from the value of the first term. We can prove this as follows, un= an2 + bn + c u1 = a × 12 + b × 1 + c u1 = a + b + c Rearranging this formula to make b the subject gives, b = u1– (a + c) If we find the value of a and c first, we can then use them and the value of the first term, to find b.

  37. The nth term of a quadratic sequence +5 +9 +13 +17 +4 +4 +4 Find the nth term of the sequence, 4, 9, 18, 31, 48, … Let’s start by looking at the first and second differences. 4, 9, 18, 31, 48 The second differences are constant and so the nth term is in the form un = an2 + bn + c. Let’s find a, b and c. The second difference is 4, so we know 2a = 4 a = 2

  38. The nth term of a quadratic sequence +4 +4 +17 +13 +4 +5 +9 +4 +1 Find the nth term of the sequence, 4, 9, 18, 31, 48, … Let’s start by looking at the first and second differences. 3, 4, 9, 18, 31, 48 The value of c is the same as the value for the 0th term. We can find this by continuing the pattern in the differences backwards from the first term. The 0th term is 3,so: c = 3

  39. The nth term of a quadratic sequence Find the nth term of the sequence, 4, 9, 18, 31, 48, … Putting a = 2 and c = 3 into un =an2 + bn + c gives us un = 2n2 + bn + 3. We can use this to write an expression for the first term: un = 2n2 + bn + 3 u1 = 2 × 12 + b × 1 + 3 u1 = 2 + b + 3 u1 = b + 5 The first term in the sequence is 4, so: 4 = b + 5 b = –1

  40. The nth term of a quadratic sequence Once we have found the values of a, b, and c we can use them in un = an2 + bn + c to give the nthterm. We have found that for the sequence 4, 9, 18, 31, 48, … a = 2, b = –1 and c = 3, un = 2n2 – n + 3 We can check this rule by substituting a chosen value for n into the formula and making sure that it corresponds to the required term in the sequence. For example, when n = 5 we have, u5 = 2 × 52 – 5 + 3 = 48 Check the rule for other terms in the sequence.

  41. The nth term of a quadratic sequence +2 +3 +4 +5 +1 +1 +1 Find the nth term of the sequence 1, 3, 6, 10, 15, … This is the sequence of triangular numbers. 1, 3, 6, 10, 15 The second differences are constant and so the nth term is in the form un =an2 + bn + c. Let’s find a, b and c. The second difference is 1, so 2a = 1 a = ½

  42. The nth term of a quadratic sequence +1 +1 +5 +4 +1 +2 +3 +1 +1 Find the nth term of the sequence 1, 3, 6, 10, 15, … This is the sequence of triangular numbers. 0, 1, 3, 6, 10, 15 The value of c is the same as the value for the 0th term. We can find this by continuing the pattern in the differences backwards from the first term. The 0th term is 0, so: c = 0

  43. The nth term of a quadratic sequence Find the nth term of the sequence 1, 3, 6, 10, 15, … Putting a = ½ and c = 0 into un =an2 + bn + c gives us un = ½n2 + bn. We can use this to write an expression for the first term: un = ½n2 + bn u1 = ½ × 12 + b× 1 u1 = ½ + b The first term in the sequence is 1, so 1 = ½ + b b = ½

  44. The nth term of a quadratic sequence n2 n un = + 2 2 or n2 + n un = 2 Once we have found the values of a, b, and c we can use them in an2 + bn + c to give the nthterm. We have found that for the sequence 1, 3, 6, 10, 15, … a = ½, b = ½ and c = 0; Checking, when n = 5, we have u5 = ½(52 + 5) = 15 Check the rule for other terms in the sequence.

  45. Finding the nth term of a quadratic sequence

  46. Tiling patterns Pattern 1 Pattern 2 1 tile Pattern 3 5 tiles Pattern 4 13 tiles 25 tiles The following patterns are made from tiles, How many tiles will there be in the next pattern? How many tiles will there be in the 20th pattern?

  47. Tiling patterns +4 +8 +12 +16 +4 +4 +4 To work out the number of tiles in the next pattern we can look at the difference between consecutive terms in the sequence, 1, 5, 13, 25, 41 If the second difference is always equal to 4 then we can use this to find the next term. Using this pattern we can predict that there will be 41 tiles in the next pattern.

  48. Tiling patterns Look at the pattern made by the differences, 1, 5, 13, 25, 41 +4 +8 +12 +16 +4 +4 +4 To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence. The second difference is constant and so the nth term will be a quadratic of the form un = an2 + bn + c. The second difference is always equal to 2a: 2a = 4 a = 2

  49. Tiling patterns +4 +4 +16 +12 +4 +4 +8 +4 +0 To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence. Look at the pattern made by the differences, 1, 1, 5, 13, 25, 41 The value of c is equal to the 0th term. The 0th term is equal to 1, so: c = 1

  50. Tiling patterns +4 +4 +16 +12 +4 +4 +8 To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence. Look at the pattern made by the differences, 1, 5, 13, 25, 41 If a = 2 and c = 1 then un = 2n2 + bn + 1, u1 = 2 + b + 1 u1 = b + 3 1 is the first term. 1 = b + 3 b = –2

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