1. Sequences and Series

2. Sequence - Is a relationship from the set of counting numbers (1, 2, 3...) to another set of numbers. Terms - The individual numbers in a sequence - Are denoted by using t followed by a number in subscript. e.g. t 1 = 1 st term, t 2 = 2 nd term nth Term - Is a general term/rule that generates numbers for a sequence - Is denoted using t n e.g. Find the first three terms of the sequences using the general terms a) t n = 2n – 1b) t n = n 2 + 3 n = 1t 1 = 2×1 – 1t 1 = 1 n = 2t 2 = 2×2 – 1t 2 = 3 n = 3t 3 = 2×3 – 1t 3 = 5 n = 1t 1 = 1 2 + 3t 1 = 4 n = 2t 2 = 2 2 + 3t 2 = 7 n = 3t 3 = 3 2 + 3t 3 = 12 Rules can also be written inside diamond brackets

3. ARITHMETIC SEQUENCES - Each term is calculated by adding or subtracting the same number each time - The number added or subtracted is called the ‘common difference’ e.g. 1, 5, 9, 13... has a common difference of 6, -1, -8, -15... has a common difference of 5 – 1 = 4 -1 – 6 = -7

4. Finding the Term - The formula for the general term (nth term) is: t n = a + (n – 1)d a = first term,d = common difference e.g. Calculate the nth term and the 27 th term of the sequence 3, 7, 11, 15... a = 3d = 7 – 3 = 4t n = 3 + (n – 1)4 t n = 3 + 4n – 4 t n = 4n – 1 t 27 = 4×27 – 1 t 27 = 107 e.g. An arithmetic sequence has t 6 = 16 and t 9 = 25. Write down the first five terms of the sequence. 16 + d + d + d = 25 16 + 3d = 25 3d = 9 d = 3 t 5 = 16 – 3 = 13 t 4 = 13 – 3 = 10 t 3 = 7 t 2 = 4 t 1 = 1 n = 27

5. e.g. Find the number of terms in the sequence 51, 47, 43, 39... 3 a = 51d = 47 – 51 = -4t n = 51 + (n – 1)-4 t n = 51 – 4n + 4 t n = 55 – 4n The find the number of terms, we need to find n or the position of the last term. 3 = 55 – 4n -52 = -4n 13 = n Therefore there are 13 terms in the sequence n = ?

6. Partial Sum of an Arithmetic Sequence - The formula to use is: a = first term,d = common difference, a = 7d = 12 – 7 = 5 n = number of terms e.g. Find the sum of the first 20 terms of the sequence 7, 12, 17, 22 n = 20 e.g. Find the general term and the sum of the sequence 8, 16, 24...120 a = 8d = 16 – 8 = 8t n = 8 + (n – 1)8 t n = 8 + 8n – 8 t n = 8n 120 = 8n 15 = nn = ?

7. Applications of Arithmetic Sequences Jane, who is preparing to run a marathon (42 km), decides that the best way to get fit is to run a little further each day until she is capable of running the marathon distance. Her running schedule is as follows: Day 1 – 1 km, Day 2 – 1.5 km, Day 3 – 2 km etc. a) Calculate how far Jane will run on Day 10 of her schedule. b) On what day of her schedule will Jane finally run the marathon distance of 42 km? c) By the time she runs the marathon distance, how far will Jane have run in total? a = 1d = 1.5 – 1 = 0.5t n = a + (n – 1)d t 10 = 1 + (10 – 1) x 0.5 t 10 = 5.5 km n = 10 a = 1d = 0.5 t n = 1 + (n – 1) x 0.5 t n = 1 + 0.5n – 0.5 t n = 0.5 – 0.5n 42 = 0.5 – 0.5n 41.5 = 0.5n 83 = n a = 1d = 0.5 n = 83 n = ?

8. GEOMETRIC SEQUENCES - Each term is calculated by multiplying the previous term by the same number each time - The number multiplied is called the ‘common ratio’ e.g. 3, 12, 48, 192... has a common ratio of 32, -16, 8, -4... has a common ratio of 12 ÷ 3 = 4 -16 ÷ 32 = -0.5

9. Finding the Term - The formula for the general term (nth term) is: t n = a × r (n – 1) a = first term,r = common ratio e.g. Calculate the nth term and the 20 th term of the sequence 6, 12, 24... a = 6r = 12 ÷ 6 = 2t n = 6 × 2 (n – 1) t 20 = 6 × 2 (20 – 1) t 20 = 3145728 e.g. Find the number of terms in the sequence: 32000, 16000, 8000...31.25 n = 20 a = 32000 r = 16000 ÷ 32000 = 0.5 t n = 32000 × 0.5 (n – 1) 31.25 = 32000 × 0.5 (n – 1) 31.25 ÷ 32000 = 0.5 (n – 1) log(31.25 ÷ 32000) = log0.5 (n – 1) log(31.25 ÷ 32000) = (n – 1) log(0.5) log(31.25 ÷ 32000) ÷ log(0.5) = (n – 1) 10 = n – 1 n = 11 Remember we need to use logs when solving index equations n = ?

10. e.g. The 4 th term of a geometric sequence is -108 and the 6 th term is -972. Find the first three terms of the sequence -108 × r × r = -972 -108 × r 2 = -972 r 2 = 9 r = 3 or -3 t 3 = -108 ÷ 3 t 2 = -36 ÷ 3 t 1 = -12 ÷ 3 = -36 t 3 = -108 ÷ -3 = 36 = -12 t 2 = 36 ÷ -3 = -12 = -4 t 1 = -12 ÷ -3 = 4

11. Partial Sum of a Geometric Sequence - The formula to use is: a = first term,r = common ratio, a = 54r = 18 ÷ 54 = 1/3 n = number of terms e.g. Find the sum of the first 12 terms of the sequence 54, 18, 6, 2... n = 12 Use brackets on the calculator correctly!

12. e.g. Find the sum of the geometric sequence 1, 2, 4...1024 a = 1r = 2 ÷ 1 = 2 t n = 1 × 2 (n – 1) 1024 = 1 × 2 (n – 1) 1024 = 2 (n – 1) log(1024) = log2 (n – 1) log(1024) = (n – 1) log(2) log(1024) ÷ log(2) = (n – 1) 10 = n – 1 n = 11 n = ?

13. Sum to Infinity of a Geometric Sequence - The formula to use is: a = first term,r = common ratio a = 48r = 12 ÷ 48 = ¼ e.g. Find the sum to infinity of the sequence 48, 12, 4... e.g. Give the first four terms of the sequence whose first term is 20 and the sum to infinity is 40 a = 20 r = ? Where -1 < r < 1 t 1 = 20 t 2 = 20 × ½ = 10 t 3 = 10 × ½ = 5 t 4 = 5 × ½ = 2.5

14. Applications of Geometric Sequences A new car is purchased from a dealer for $32000. Each year it loses 20% of its previous years value. a) Give the first 3 terms of the sequence representing the car’s worth after 1, 2 and 3 years. b) What is the car’s worth (value) after 5 years? c) How much has the car lost in value after 10 years? r = 1 – 0.2 = 0.8 t 1 = 32000 × 0.8 = $25600 a = 25600r = 0.8 n = 10 t 2 = 25600 × 0.8 = $20480 t 3 = 20480 × 0.8 = $16384 a = 25600r = 0.8 t n = 25600 × 0.8 (n – 1) t 5 = 25600 × 0.8 (5 – 1) t 5 = $10485.76n = 5 t n = a × r (n – 1) t n = 25600 × 0.8 (n – 1) t 10 = 25600 × 0.8 (10 – 1) t 10 = $3435.97 t n = a × r (n – 1) Value lost = 32000 – 3435.97 = $28564.03 t n = a × r (n – 1)

15. d) The owner of the car decides to keep is until it is worth no less the $5000. What is the maximum number of years he can keep the car? a = 25600 r = 0.8 t n = 25600 × 0.8 (n – 1) 5000 = 25600 × 0.8 (n – 1) 5000 ÷ 25600 = 0.8 (n – 1) log(5000 ÷ 25600) = log0.8 (n – 1) log(5000 ÷ 25600) = (n – 1) log(0.8) log(5000 ÷ 25600) ÷ log(0.8) = (n – 1) 7.3 = n – 1 n = 8.3 n = ? Maximum number of years = 8 Hints for Application Questions 1. Firstly check if sequence is arithmetic or geometric 2. Determine if it is a term you are seeking or a sum and choose formula 3. List all of the information known to help decide what you need to find. 4. Substitute information and solve as appropriate. Remember: If unsure you can always Guess and Check.

16. Using Simultaneous Equations - Sometimes you may need to use simultaneous equations when working with two separate sequences. One type of example: When will the following sequences have the same total? Sequence A = 17, 23, 29, 35... Sequence B = 12, 19, 26, 33... - The equations may involve the formulas for terms or sums of both the arithmetic and geometric sequences. a = 17d = 23 – 17 = 6n = ? a = 12d = 19 – 12 = 7n = ? Therefore both sequences will have the same total when n = 11

17. Sigma Notation - Where the symbol Σ (sigma) is used to mean ‘the sum of’ - The numbers below and above the symbol show the first and last numbers in the sequence to be added. e.g. For the following sequence: 4, 7, 10, 13, 16... evaluate t 4 = 13t 5 = 16t 6 = 19t 7 = 22 = 70 Evaluate the following: = (3×1 + 1) + (3×2 + 1) + (3×3 + 1) + (3×4 + 1) = 4 + 7 + 10 + 13 = 34 = (3×1) + (3×2) + (3×3) + (3×4) + 1 = 3 + 6 + 9 + 12 + 1 = 31 = (4×1 2 ) + (4×2 2 ) + (4×3 2 ) + (4×4 2 ) = 4 + 16 + 36 + 64 = 120

18. Express the following sequences in sigma notation: a) 51, 47, 43... 3 b) 1, 2, 4... 1024 a = 51d = 47 – 51 = -4 t n = 51 + (n – 1)×-4 t n = 51 – 4n + 4 t n = -4n + 55 t n = a + (n – 1)d 3 = 55 – 4n -52 = -4n 13 = n First find nth term Then find n of last term a = 1r = 2 ÷ 1 = 2 t n = a × r (n – 1) t n = 1 × 2 (n – 1) 1024 = 1 × 2 (n – 1) 1024 = 2 (n – 1) log(1024) = log2 (n – 1) log(1024) = (n – 1) log(2) log(1024) ÷ log(2) = (n – 1) 10 = n – 1 n = 11 First find nth term Then find n of last term

19. Recursive Definition - When the first term t 1 is given - The formula tells us how to get the next term t n +1 from the term before t n e.g. t 1 = 4, t n + 1 = 3t n – 2. Write down the first four terms of the sequence t 1 = 4t 2 = t 1 + 1 t 2 = 3t 1 – 2 t 2 = 3×4 – 2 t 2 = 10 t 3 = t 2 + 1 t 3 = 3t 2 – 2 t 3 = 3×10 – 2 t 3 = 28 t 4 = t 3 + 1 t 4 = 3t 3 – 2 t 4 = 3×28 – 2 t 4 = 82 e.g. t 1 = -1, t n + 1 = 2(t n ) 2 + 5. Write down the first four terms of the sequence. t 1 = -1t 2 = t 1 + 1 t 2 = 2(t 1 ) 2 + 5 t 2 = 2(-1) 2 + 5 t 2 = 7 t 3 = t 2 + 1 t 3 = 2(t 2 ) 2 + 5 t 3 = 2(7) 2 + 5 t 3 = 103 t 4 = t 3 + 1 t 4 = 2(t 3 ) 2 + 5 t 4 = 2(103) 2 + 5 t 4 = 21223

20. e.g. t 1 = 2, t 2 = 5, and t n + 1 = t n – 2t n – 1. Write down the first four terms of the sequence t 1 = 2t 2 = 5t 3 = t 2 + 1 t 3 = t 2 – 2t 1 t 3 = 5 – 2×2 t 3 = 1 t 4 = t 3 + 1 t 4 = t 3 – 2t 2 t 4 = 1 – 2×5 t 4 = -9

21. REAL LIFE EXAMPLES OF GEOMETRIC SEQUENCES

22. Compound Interest - Where interest is paid on interest. - A formula that can be used is: A = new principal P = original principal r = interest rate n = number of years e.g. a) After investing for 10 years at an interest rate of 8% a person has a total of $45000. What was their initial investment? b) After investing $15000 for 12 years, the value of the investment is $26000. What was the annual compound interest rate? A = 45000r = 8 n = 10P = ? A = 26000r = ? n = 12P = 15000

23. Inflation - Where the value of an item increases. - A formula that can be used is: I n = Increased price O n = Old price r = inflation rate n = number of years e.g. a) An antique chest initially purchased for $16000 appreciates at a rate of 8.2% per year. What is its value in 10 years time? b) A piece of land originally brought for $10000 has been valued at $85000. If the rate of inflation was 24%, how many years ago was it purchased? I = ?r = 8.2 n = 10O = 16000 I = 85000r = 24 n = ?O = 10000

24. Depreciation - Where the value of an item decreases. - A formula that can be used is: D n = Decreased price O n = Old price r = depreciation rate n = number of years e.g. a) A laptop was originally purchased for $3500 and depreciates at a rate of 26% per annum. What is its value in 3 years time? b) A car was purchased in 1998 for $5900. In 2002 its value was $500. What has been the yearly percentage loss in value? D = ?r = 26 n = 3O = 3500 D = 500r = ? n = 4O = 5900

25. - Depends on how much of substance is initially present. - A formula that can be used is: D n = Amount left O n = Original amount r = rate of decay n = number of years e.g. A radioactive substance has a half life of 26 years. If we initially start with 500 g a) What is the rate of decay of the substance? D = 250r = ? n = 26O = 500 Radioactive Decay - How life = length of time it takes material to reduce to half its previous value.

26. b) How much of the substance is left after 20 years? D = ?r = 2.6 n = 20O = 500 c) How many years does it take for the substance to decay down to 150 g? D = 150r = 2.6 n = ?O = 500