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SEQUENCES Unit 2 General Maths DESCRIBING SEQUENCES This topic investigates different types of patterns and how they can be manipulated mathematically. The dictionary describes a sequence as, ‘a number of things, actions, or events arranged or happening in a specific order or having a specific connection’. In maths, the term sequence is used to represent an ordered set of elements. In this topic we will examine the relationships and patterns of these sets of data. RECOGNISING PATTERNS For these examples: eg1) 2, 4, 6, 8…….. Describe the pattern in words; Describe the pattern in mathematical terms; State the next 3 numbers in the pattern. Increasing by 2 +2 10, 12, 14 Doubling each number x2 80, 160, 320 eg2) 5, 10, 20, 40….. eg3) 1000, 500, 250…… Halving each number ÷2 or × 0.5 125, 62.5, 31.25 USING A RULE TO GENERATE A NUMBER PATTERN For these examples: eg1) Use the following rules to write down the first five numbers of each number pattern Start with a 4 and multiply by 2 each time. 4, 8, 16, 32, 64 GEOMETRIC SEQUENCE – changes by a ‘common ratio’ eg2) Start with 21 and add 6 each time. 21, 27, 33, 39, 45 ARITHMETIC SEQUENCE – changes by a ‘common difference’ ARITHMETIC SEQUENCES We can label the first term in the sequence ′𝑎′ We can label the ‘common difference’ between consecutive terms Label the arithmetic sequences with their ‘a’ and ‘d’ values 𝑎=2 eg. 2, 3, 4, 5, 6, 7, 8, 9, 10 eg. 0, 3, 6, 9, 12, 15, 18, 21, 24 eg. 4, 6, 8, 10, 12, 14 𝑑=1 𝑎=0 𝑎=4 𝑑=3 𝑑=2 ′𝑑′ ARITHMETIC SEQUENCES An ARITHMETIC SEQUENCE is one in which the difference between any two consecutive terms is the same. Are these Arithmetic Sequences? eg. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 Yes – common difference eg. 0, 3, 6, 9, 12, 15, 18, 21, 24 Yes – common difference eg. 2, 4, 5, 10, 11, 16 No – the difference between consecutive terms is NOT the same eg. 2, 4, 8, 16, 32, 64 No – the difference between consecutive terms is NOT the same ARITHMETIC SEQUENCES The terms at each position 𝑡𝑛 , in order can be labelled 𝑡1 , 𝑡2 , 𝑡3 , … … 𝑡𝑛 Label the terms in the following examples: eg. 2, 3, 4, 5, 6, 7, 8, 9, 10 𝑡1 = 2, 𝑡2 = 3, 𝑡3 = 4 𝑒𝑡𝑐 … eg. 0, 3, 6, 9, 12, 15, 18, 21, 24 𝑡1 = 0, 𝑡2 = 3, 𝑡3 = 6 𝑒𝑡𝑐 … eg. 𝑡1 = 4, 𝑡2 = 6, 𝑡3 = 8 𝑒𝑡𝑐 … 4, 6, 8, 10, 12, 14 FINDING TERMS OF AN ARITHMETIC SEQUENCE If we know the first term 𝑎, and the common difference 𝑑, we can find any number within the arithmetic sequence. The rule for finding a term in an arithmetic sequence is: The common difference 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 The value of the nth term (the term we are trying to find) The first term in the sequence The position of the term we are trying to find in the sequence Rule for Arithmetic Sequences The value of the n th term (the term we are trying to find) eg1. Consider the arithmetic sequence 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 The first term in the sequence The common difference The position of the term we are trying to find in the sequence 22, 28, 34, 40, ….. a) Write a rule for the arithmetic sequence 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 𝑡𝑛 = 22 + 6 𝑛 − 1 b) What number would be at the 30th position? 𝑙𝑒𝑡 𝑛 = 30: 𝑡𝑛 = 22 + 6 𝑛 − 1 𝑡30 = 22 + 6 30 − 1 𝑡30 = 22 + 6 29 = 22 + 174 = 196 Rule for Arithmetic Sequences 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 eg2. Consider the arithmetic sequence 10, 13, 16, 19, 22, 25,… a) Write a rule for the arithmetic sequence 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 𝑡𝑛 = 10 + 3 𝑛 − 1 b) What number would be at the 16th position? 𝑡𝑛 = 10 + 3 𝑛 − 1 𝑙𝑒𝑡 𝑛 = 16: 𝑡16 = 10 + 3 16 − 1 𝑡16 = 10 + 3 15 = 10 + 45 = 55 Rule for Arithmetic Sequences 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 eg3. Consider the arithmetic sequence 41, 37, 33, 29, 25,…. a) Write a rule for the arithmetic sequence 𝑡𝑛 = 41 − 4 𝑛 − 1 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 b) What number would be at the 11th position? 𝑙𝑒𝑡 𝑛 = 30: 𝑡𝑛 = 41 − 4 𝑛 − 1 𝑡30 = 41 − 4 11 − 1 𝑡30 = 41 − 4 10 = 41 − 40 = 1 Rule for Arithmetic Sequences 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 eg4. Consider a sequence which starts at 14 and each number increases by 4.5. Find the first 5 numbers in the sequence. 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 𝑡𝑛 = 14 + 4.5 𝑛 − 1 𝑙𝑒𝑡 𝑛 = 2: 𝑙𝑒𝑡 𝑛 = 3: 𝑡2 = 14 + 4.5 2 − 1 𝑡2 = 14 + 4.5 1 = 18.5 𝑡3 = 14 + 4.5 3 − 1 𝑡3 = 14 + 4.5 2 = 14 + 9 = 23 etc…. OR use your calculator Rule for Arithmetic Sequences 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 Eg4 (continued). Consider a sequence which starts at 14 and each number increases by 4.5. Find the first 5 numbers in the sequence. 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 𝑡𝑛 = 14 + 4.5 𝑛 − 1 OR use your calculator in another way.. 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 Eg4 (continued) Using the SEQUENCES function. Choose Explicit 𝑡𝑛 = 14 + 4.5 𝑛 − 1 Change these values for ‘n’ to produce a table of values as small or large as you require NOW DO Exercise 6.2 Q1, 2, 3abc, 4abc, 5, 6, 7 FINDING OTHER VARIABLES IN AN ARITHMETIC SEQUENCE We can rearrange the rule to find a, d or n: 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 Finding the first number in a sequence: Finding the common difference: Finding the position of the term: 𝑑= 𝑎 = 𝑡𝑛 − 𝑑 𝑛 − 1 𝑡𝑛 − 𝑎 𝑛−1 𝑛= 𝑡𝑛 − 𝑎 𝑑 +1 NOW DO Exercise 6.2 Q1, 2, 3abc, 4abc, 5, 6, 7, 8abc, 15ac, 16ac, 17a GEOMETRIC SEQUENCES A geometric sequence is one where the next term is formed by multiplying the previous term by a fixed number called the common ratio, 𝑟 The next term in the sequence is found by multiplying the current term by the common ratio, giving the equation: 𝑡𝑛+1 = 𝑟 𝑡𝑛 Current Term Next term Common Ratio Rearranging this equation allows us to find the common ratio, using: 𝑡𝑛+1 𝑟= 𝑡𝑛 Next term Current Term GEOMETRIC SEQUENCES We can decide if a sequence is a geometric sequence by calculating the ratio for subsequent terms and checking that the ratio is consistent throughout the sequence. 𝑡𝑛+1 𝑟= 𝑡𝑛 Next term Current Term eg. Are the following sequences geometric? a) 5, 10, 20, 40, 80…. YES 𝑟= 𝑡𝑛+1 𝑡𝑛 𝑟= 𝑡𝑛+1 𝑡𝑛 = 10 5 = 20 10 =2 =2 𝑟= 𝑡𝑛+1 𝑡𝑛 𝑟= 𝑡𝑛+1 𝑡𝑛 = 40 20 =2 = 80 40 =2 GEOMETRIC SEQUENCES We can decide if a sequence is a geometric sequence by calculating the ratio for subsequent terms and checking that the ratio is consistent throughout the sequence. 𝑡𝑛+1 𝑟= 𝑡𝑛 Next term Current Term eg. Are the following sequences geometric? a) 100, 50, 25, 12.5…. YES 𝑟= 𝑡𝑛+1 𝑡𝑛 𝑟= 𝑡𝑛+1 𝑡𝑛 = 50 100 = 25 50 = 0.5 = 0.5 𝑟= 𝑡𝑛+1 𝑡𝑛 𝑟= 𝑡𝑛+1 𝑡𝑛 = 25 50 = 12.5 25 = 0.5 = 0.5 GEOMETRIC SEQUENCES We can decide if a sequence is a geometric sequence by calculating the ratio for subsequent terms and checking that the ratio is consistent throughout the sequence. 𝑡𝑛+1 𝑟= 𝑡𝑛 Next term Current Term eg. Are the following sequences geometric? a) 2, 10, 20, 60…. NO 𝑟= 𝑡𝑛+1 𝑡𝑛 𝑟= 𝑡𝑛+1 𝑡𝑛 = 10 2 =5 = 20 10 =2 𝑟= 𝑡𝑛+1 𝑡𝑛 = 60 20 =3 GEOMETRIC SEQUENCE A geometric sequence can be written in terms of its 1st term ‘a’, and it’s common ratio ‘r’ 𝑎, 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ………..etc 1st 2nd 3rd Term Term Term Any term n, in the sequence, can be found if we know the 1st term and the ratio, using: 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 eg. A geometric sequence begins with 6 and has a common ratio of 4. Find the 3rd term using the rule above. 𝑡𝑛 = 𝑎𝑟 𝑛−1 𝑡3 = 6 × 43−1 = 6 × 42 = 6 × 16 = 96 Lets prove it using our calculators.. Finding the 𝑛𝑡ℎ term 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 𝑎, eg. For the sequence 4, 8, 16, 32.. Find the 10th term of the sequence. 𝑡𝑛 = 𝑎𝑟 𝑛−1 𝑡10 = 4 × 210−1 = 4 × 29 = 4 × 512 = 2048 1st Term 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ……..etc 2nd Term 3rd Term Finding the 𝑛𝑡ℎ term 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 𝑎, eg. For the sequence 80, 40, 20, 10, 5.. Find the 8th term of the sequence. 𝑡𝑛 = 𝑎𝑟 𝑛−1 𝑡8 = 80 × (0.5)8−1 = 80 × (0.5)7 = 80 × 0.0078125 = 0.625 1st Term 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ……..etc 2nd Term 3rd Term NOW DO APPLICATIONS PROBLEMS In this exercise, we look at solving real life, worded problems using the arithmetic and geometric sequences and series equations from all prior exercises. 1. Read the question carefully 2. Decide whether the problem involves an arithmetic or geometric sequence 3. Write the information given, in terms of variables (ie. 𝑎, 𝑑, 𝑟, 𝑡𝑛 , 𝑒𝑡𝑐.) 4. Choose the appropriate formula to use to solve the problem. 5. Solve and answer the question being asked in the problem. EQUATIONS Arithmetic Sequences and Series Geometric Sequences and Series Finding the nth term Finding the nth term 𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 APPLICATIONS PROBLEMS If $1000 is put into a compound interest account, at an interest rate of 8%, how much will be in the account after 10 years? geometric sequence 𝒂 = 𝟏𝟎𝟎𝟎 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 10−1 𝑡10 = 1000 × (1.08) 𝒓 = 𝟏. 𝟎𝟖 𝑡10 = 1000 × (1.08)9 𝒏 = 𝟏𝟎 𝑡10 = 1000 × 1.999004627 = 1999.004627 After 10 years, the account balance will be $1999.00 APPLICATIONS PROBLEMS The seating arrangements at a concert are: first row has 10 seats, second row has 13 seats, third row has 16 seats, following this pattern all the way through to the final 20th row. a) How many seats are in the 20th row? b) How many seats are there in total? 𝒂 = 𝟏𝟎 𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑 𝑡20 = 10 + 20 − 1 3 𝐝=𝟑 𝑡20 = 10 + 19 × 3 Arithmetic sequence 𝒏 = 𝟐𝟎 = 10 + 57 = 67 𝑠𝑒𝑎𝑡𝑠 APPLICATIONS PROBLEMS The seating arrangements at a concert are: first row has 10 seats, second row has 13 seats, third row has 16 seats, following this pattern all the way through to the final 20th row. a) How many seats are in the 20th row? b) How many seats are there in total? 𝑛 𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑) Arithmetic sequence 2 𝒂 = 𝟏𝟎 𝐝=𝟑 𝒏 = 𝟐𝟎 𝑆20 20 = ( (2 × 10) + 20 − 1 3) 2 𝑆20 = 10 19 × 3 = 10 × 57 = 570 NOW DO Use 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 to form equations for each given term Now use the simultaneous equation solver on your calculator to solve for our unknowns r and a Term 3 = 100 𝑡3 = 𝑎 × 𝑟 3−1 100 = 𝑎𝑟 2 Term 5 = 400 𝑡5 = 𝑎 × 𝑟 400 = 𝑎𝑟 4 5−1 So the first term, a = 25 The common ratio is 2 Use 𝑡𝑛 = 𝑎 × 𝑟 𝑛−1 to form equations for each given term Now use the simultaneous equation solver on your calculator to solve for our unknowns r and a Term 4 = 90 𝑡4 = 𝑎 × 𝑟 4−1 90 = 𝑎𝑟 3 So the first term, a = 3.33 Term 7 = 2430 The common ratio is 3 𝑡7 = 𝑎 × 𝑟 2430 = 𝑎𝑟 6 7−1 COMBINED PROBLEMS Problems involving both Arithmetic and Geometric Sequences Some sequences involve both a common difference and a common ratio. These take the form: 𝒕𝒏+𝟏 = 𝒓 × 𝒕𝒏 + 𝒅, 𝒕𝟏 = 𝒂 Consider the sequence, ‘Start with 3, Multiply the number by 2 and add 4. The recurrence relation is: 𝒕𝒏+𝟏 = 𝟐𝒕𝒏 + 𝟒, 𝒕𝟏 = 𝟑 Try these for yourself on your worksheet ARITHMETIC SEQUENCES GENERATED BY RECURSION A sequence can be generated by the repeated use of an instruction. This is known as recursion. We can form equations to model recursion: The current term given by 𝑛 can be represented by 𝑡𝑛 The following term is 𝑡𝑛+1 & the term before it is 𝑡𝑛−1 Using this, we can form an equation to model the recursion. The recursion relation is: 𝑡𝑛+1 = 𝑡𝑛 + 𝑑 , The next term The current term 𝑡1 = 𝑎 The common difference The first term in the sequence Recursion relation The next term 𝑡𝑛+1 = 𝑡𝑛 + 𝑑, The current term 𝑡1 = 𝑎 The common difference Form recursion relations for the following: eg. 2, 3, 4, 5, 6, 7, 8, 9, 10 𝑡𝑛+1 = 𝑡𝑛 + 1, 𝑡1 = 2 eg. 0, 3, 6, 9, 12, 15, 18, 21, 24 𝑡𝑛+1 = 𝑡𝑛 + 3, 𝑡1 = 0 eg. 𝑡𝑛+1 = 𝑡𝑛 + 0.5, 𝑡1 = 4 𝑡𝑛+1 = 𝑡𝑛 − 3, 𝑡1 = 30 4, 4.5, 5, 5.5, 6, 6.5 eg. 30, 27, 24, 21, 18, 15 The first term in the sequence NOW DO Chapter 5 Recursion Worksheet – Exercise 1 GEOMETRIC SEQUENCES WRITING A RECURRENCE RELATION So, we can write the recurrence relation for geometric sequences: 𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎 1st Term Common Ratio eg. Given the sequence 1, 2, 4, 8, 16, 32…… a) The sequence could be described as: Start at 1 and multiply by 2 for subsequent terms b) Form the recurrence relation for the sequence: c) Its this growth or decay? Geometric Growth 𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎 𝑡𝑛+1 = 2𝑡𝑛 , 𝑡1 = 1 𝑡𝑛+1 = 𝑟 𝑡𝑛 , recurrence relation Common Ratio eg. Given the sequence 𝑡1 = 𝑎 1st Term 2, 6, 18, 54, 162… a) The sequence could be described as: Start at 2 and multiply by 3 for subsequent terms b) Form the recurrence relation for the sequence: c) Its this growth or decay? Geometric Growth 𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎 𝑡𝑛+1 = 3𝑡𝑛 , 𝑡1 = 2 𝑡𝑛+1 = 𝑟 𝑡𝑛 , recurrence relation Common Ratio eg. Given the sequence 𝑡1 = 𝑎 1st Term 100, 50, 25, 12.5… a) The sequence could be described as: Start at 100 and divide by 2 for subsequent terms b) Form the recurrence relation for the sequence: 𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎 c) Its this growth or decay? Geometric Decay 𝑡𝑛+1 = 1 𝑡 2 𝑛 , 𝑡1 = 100 NOW DO Recursion Worksheet Part 2