1. HW: Do Now Aim : How do we write the terms of a Sequence? Write the first 5 terms of the sequence, then find the thirty-first term:

2. ARITHMETIC SEQUENCES the difference between successive terms of a sequence is always the same number. are sequences where … This number is called the common difference.

3. 3, 7, 11, 15, 19 … 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. d = 4 To get to the next term in the sequence we would add 4 so a recursive formula for this sequence is: The first term in the sequence would be a 1 which is sometimes just written as a. a = 3

4. 3, 7, 11, 15, 19 … +4 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. d = 4 Try this to get the 5th term. a = 3

5. Let’s look at a formula for an arithmetic sequence and see what it tells us. Subbing in the set of positive integers we get: 3, 7, 11, 15, 19, … What is the common difference? d = 4 you can see what the common difference will be in the formula We can think of this as a “compensating term”. Without it the sequence would start at 4 but this gets it started where we want it. 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,...?

6. 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. Check it out by putting in the first few positive integers and verifying that it generates our sequence. 6, 4, 2, 0, -2,... Sure enough---it starts at 6 and has a common difference of -2

7. Finding Terms of an Arithmetic Sequence Find the first six terms and the 300th term of the arithmetic sequence 13, 7,... Since the first term is 13, we have a = 13. The common difference is: d = 7 – 13 = –6 Thus, the nth term of the sequence is: a n = 13 – 6(n – 1)

8. From that, we find the first six terms: 13, 7, 1, –5, –11, –17,... The 300th term is: a 300 = 13 – 6(299) = –1781

9. 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 you add to get from the 4th term to the 20th term? 353 Solve this for d d = 2 The fourth term is the first term plus 3 common differences. 3 (2) We have all the info we need to express these sequences. We’ll do it on next slide.

10. 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. d = 2 makes the common difference 2 makes the first term - 3 instead of 2 The recursive formula would be: Let’s check it out. If we find n = 4 we should get the 4th term and n = 20 should generate the 20th term.

11. Practice The 11th term of an arithmetic sequence is 52, and the 19 th term is 92. Find the 1000 th term.

12. Since a 11 = 52 and a 19 = 92, we get the two equations: Solving this for a and d, we get: a = 2, d = 5 The nth term is: a n = 2 + 5(n – 1) The 1000th term is: a 1000 = 2 + 5(999) = 4997

13. Arithmetic Means 17, 10, 3, -4, -11, -18, … We say that, between 10 and -18 there are three arithmetic means 3, -4, -11 Problem : Find three arithmetic means between 8 and 14.

14. So our sequence must look like 8, __, __, __, 14. In order to find the means we need to know the common difference. We can use our formula to find it. Problem : Find three arithmetic means between 8 and 14.

15. 8, __, __, __, 14 a1 = 8 a5 = 14 n = 5 14 = 8 + d(5 - 1) 14 = 8 + d(4)subtract 8 6 = 4ddivide by 4 1.5 = d Problem : Find three arithmetic means between 8 and 14.

16. 8, __, __, __, 14 so to find our means we just add 1.5 starting with 8. Answer : 8, 9.5, 11, 12.5, 14

17. Additional Example 72 is the __ term of the sequence -5, 2, 9, … We need to find ‘n’ which is the term number. 72 is a n, -5 is a 1, 7 is d. Plug it in.

18. Additional Example 72 = -5 + 7(n - 1) 72 = -5 + 7n - 7 72 = -12 + 7n 84 = 7n n = 12 72 is the 12th term.

19. Find 5 arithmetic means between 2 and 23.