1. ARITHMETIC SEQUENCES AND SERIES Week Commencing Monday 28 th September Learning Intention: To be able to find the nth term of an arithmetic sequence or series. To be able to find the number of terms in an arithmetic sequence or series. Contents: 1.What is an Arithmetic Sequence?What is an Arithmetic Sequence? 2.What is an Arithmetic Series?What is an Arithmetic Series? 3.Assignment 2Assignment 2 4.Finding terms of Arithmetic Sequences and SeriesFinding terms of Arithmetic Sequences and Series 5.Number of terms in a Sequence or SeriesNumber of terms in a Sequence or Series 6.Finding first term and common differenceFinding first term and common difference 7.Assignment 3Assignment 3

2. ARITHMETIC SEQUENCES AND SERIES What is an Arithmetic Sequence? An arithmetic sequence is a sequence that increases by a constant amount each time. It can be defined by the recurrence relationship: U n+1 = U n + k, where k is a constant number Examples of arithmetic sequences are: 5, 8, 11, 14, 17,... increasing by 3 each time 100, 95, 90, 85,... increasing by -5 each time

3. ARITHMETIC SEQUENCES AND SERIES What is an Arithmetic Series? If you add together the terms of an arithmetic sequence we get an arithmetic series – the same terms but instead of comma’s separating them it is a “+” sign. Examples of arithmetic series are: 5 + 8 + 11 + 14 + … 5 + 1 + -3 + -7 + -11 + …

4. ARITHMETIC SEQUENCES AND SERIES Assignment 2 – What are Arithmetic Sequences & Series? Follow the link for Assignment 2 on Arithmetic Sequences and Series in the Moodle Course Area. Completed assignments must be submitted by 5:00pm on Monday 5 th October.

5. ARITHMETIC SEQUENCES AND SERIES Finding terms of Arithmetic Sequences and Series For both arithmetic sequences and series the first term is generally called a and the constant it increases by is called the common difference, d. We can use a and d to help us find the nth term of an arithmetic sequence or series. The formula for the nth term is given by: a + (n – 1)d where n is term we are looking for a is the first term d is the common difference

6. ARITHMETIC SEQUENCES AND SERIES Terms of an Arithmetic Series Example: Find the 10 th, 20 th and nth terms of this arithmetic series: ¼ + 1 + 1¾ + 2½ + … Solution: a = ¼ d = 1 – ¼ = ¾ Using a + (n -1)d (i)10 th term = ¼ + (10 – 1)¾ = ¼ + (9)¾ = 7 a = ¼ d = 1 – ¼ = ¾ Using a + (n -1)d (ii) 20 th term = ¼ + (20 – 1)¾ = ¼ + (19)¾ = 14½ a = ¼ d = 1 – ¼ = ¾ Again, using a + (n -1)d (iii) n th term = ¼ + (n – 1)¾ = ¼ + ¾n – ¾ = ¾n – ½

7. ARITHMETIC SEQUENCES AND SERIES Number of Terms in a Series If we know the final term in a sequence or series we can use a and d to help us find how many terms there are in sequence or series.

8. ARITHMETIC SEQUENCES AND SERIES Number of Terms in a Series? Example: How many terms are in this arithmetic series: 0.7 + 0.3 + -0.1 + -0.5 + … + -5.7 Solution: We know the last term is -5.7, a = 0.7 and d = -0.4. We can therefore use the formula a + (n – 1)d to form an equation and solve for n. We get: 0.7 + ( n – 1)(-0.4) = -5.7 0.7 – 0.4n + 0.4 = -5.7 (multiplying out brackets) -0.4n = -6.8 (taking numbers to one side) n = -6.8 / -0.4 = 17 (dividing by 0.4)

9. ARITHMETIC SEQUENCES AND SERIES Finding a and d A very popular type of question to be asked in the exam is to find the first term and the common difference when given what two of the terms in the series are.

10. ARITHMETIC SEQUENCES AND SERIES Finding a and d Example: The seventh term in an arithmetic series is 15 and the eight term is 20. Find the first term. Solution: U 7 = 15 and U 8 = 20, therefore d = 5. Furthermore: a + (7 -1)(5) = 15 a + 30 = 15 a = 15 – 30 = -15

11. ARITHMETIC SEQUENCES AND SERIES Finding a and d Example: Given that the 3 rd term of an arithmetic series is 30 and the 10 th term is 9 find a and d. Hence find which term if the first one to become negative. Solution: U 3 = 30 and U 10 = 9 a + (3 -1)d = 30a + (10 – 1)d = 9 a + 2d = 30(1)a + 9d = 9(2) We solve equations (1) and (2) simultaneously to find a and d. Subtracting (1) from (2) gives: 7d = -21 d = -3 Therefore, a + 2(-3) = 30 a – 6 = 30 a = 36 a = 36d = -3 We want the first term to become negative i.e a + (n – 1)d < 0 Using the a and d we have found we get: 36 + (n – 1)(-3) < 0 36 – 3n + 3 < 0 -3n < -39 n > 13 That is, from term number 14 onwards the number will be negative.

12. ARITHMETIC SEQUENCES AND SERIES Assignment 3 – Finding terms of an Arithmetic Series. Follow the link for Assignment 3 on Finding terms of an Arithmetic Series in the Moodle Course Area. This is a Yacapaca Activity. Completed assignments must be submitted by 5:00pm on Monday 5 th October.