VIEWS: 266 PAGES: 11 POSTED ON: 5/6/2010 Public Domain
ARITHMETIC SEQUENCES These are sequences where the difference between successive terms of a sequence is always the same number. This number is called the common difference. 3, 7, 11, 15, 19 … d=4 a=3 Notice in this sequence that if we find the difference between any term and the term before it we always get 4. 4 is then called the common difference and is denoted with the letter d. To get to the next term in the sequence we would add 4 so a recursive formula for this sequence is: an an 1 4 The first term in the sequence would be a1 which is sometimes just written as a. +4 +4 +4 +4 d=4 a=3 3, 7, 11, 15, 19 … Each time you want another term in the sequence you’d add d. This would mean the second term was the first term plus d. The third term is the first term plus d plus d (added twice). The fourth term is the first term plus d plus d plus d (added three times). So you can see to get the nth term we’d take the first term and add d (n - 1) times. an a n 1d Try this to get the 5th term. a5 3 5 14 3 16 19 Let’s look at a formula for an arithmetic sequence and see what it tells us. 4n 1 you can see what the We can think of this as common difference a “compensating term”. will be in the formula Without it the sequence would start at 4 but this gets it started where we want it. Subbing in the set of positive integers we get: What is the 3, 7, 11, 15, 19, … common d=4 difference? 4n 2 4n would generate the multiples of 4. With the - 1 on the end, everything is back one. What would you do if you wanted the sequence 2, 6, 10, 14, 18, . . .? Find the nth term of the arithmetic sequence when a = 6 and d = -2 If we use -2n we will generate a sequence whose common difference is -2, but this sequence starts at -2 (put 1 in for n to get first term to see this). We want ours to start at 6. We then need the “compensating term”. If we are at -2 but want 6, we’d need to add 8. 2n 8 Check it out by putting in the first few positive integers and verifying that it generates our sequence. Sure enough---it starts at 6 and 6, 4, 2, 0, -2, . . . has a common difference of -2 Let’s try something a little trickier. What if we just know a couple of terms and they aren’t consecutive? The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence. How many differences would a 3, a 35 4 20 you add to get from the 4th term to the 20th term? a20 a4 16 d 35 3 Solve this for d d=2 3 4 a1 3d The fourth term is the first term plus 3 common differences. a (2) a1 3 We have all the info we need to express these sequences. We’ll do it on next slide. The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence. a4 3, a20 35 d=2 a1 3 makes the first term - 3 instead of 2 a4 24 5 3 2n 5 a20 220 5 35 makes the common difference 2 Let’s check it out. If we find n = 4 we should get the 4th term and n = 20 should generate the 20th term. The recursive formula would be: an an 1 2 Often in applications we will want the sum of a certain number of terms in an arithmetic sequence. The story is told of a grade school teacher In the 1700's that wanted to keep her class busy while she graded papers so she asked them to add up all of the numbers from 1 to 100. These numbers are an arithmetic sequence with common difference 1. Carl Friedrich Gauss was in the class and had the answer in a minute or two (remember no calculators in those days). This is what he did: sum is 101 1 + 2 + 3 + 4 + 5 + . . . + 96 + 97 + 98 + 99 + 100 sum is 101 With 100 numbers there are 50 pairs that add up to 101. 50(101) = 5050 This will always work with an arithmetic sequence. The formula for the sum of n terms is: n is the number of terms so n/2 would be the number of pairs n 2 S n a1 an first term last term Let’s find the sum of 1 + 3 +5 + . . . + 59 But how many terms are there? We can write a formula for the sequence and then figure out what term number 59 is. n 2 S n a1 an first term last term Let’s find the sum of 1 + 3 +5 + . . . + 59 The common difference is 2 and the first term is one so: 2n 1 Set this equal to 59 to find n. Remember n is the term number. 2n - 1 = 59 n = 30 So there are 30 terms to sum up. S30 1 59 900 30 2 Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au