Notes 12.1 Arithmetic Sequences and Series.pdf
- 1. I. Sequences and Terms • Sequence: a list of numbers in a specific order. 1, 3, 4, 7, 10, 16 • Term: each number in a sequence Sequence Terms Notes 12.1: Arithmetic Sequences and Series
- 2. A sequence is arithmetic if the differences between consecutive terms are the same. 4, 9, 14, 19, 24, . . . 9 – 4 = 5 14 – 9 = 5 19 – 14 = 5 24 – 19 = 5 arithmetic sequence The common difference, d, is 5. II. Arithmetic Sequences FYI: Common differences can be negative. A. Definition
- 3. How do I know if it is an arithmetic sequence? • Look for a common difference between consecutive terms Ex: 2, 4, 8, 16 . . . Common Difference? 2+2 = 4 4+ 4 = 8 8 + 8 = 16 No. The sequence is NOT Arithmetic Ex: 48, 45, 42, 39 . . . Common Difference? 48 - 3 = 45 45 - 3 = 42 42 - 3 = 39 Yes. The sequence IS Arithmetic. d = -3
- 4. III. Finding Subsequent Terms • Find the next three terms in the arithmetic sequence: 2, 5, 8, 11, 14, __, __, __ • 2, 5, 8, 11, 14, 17, 20, 23 • The common difference is? • 3!!!
- 5. IV. Finding the nth Term of an Arithmetic Sequence an = a1 + (n - 1)d Where: an is the nth term in the sequence a1 is the first term n is the number of the term d is the common difference
- 6. Ex 1: Find the 25th term in the sequence of 5, 11, 17, 23, 29 . . . an = a1 + (n - 1)d a25 = 5 + (24)6 =149 Common difference a2 - a1 = 11 – 5 = 6 a25 = 5 + (25 -1)6 Start with the explicit sequence formula Find the common difference between the values. Substitute in known values Simplify
- 7. Ex 2: Find the 17th term of the arithmetic sequence: 26, 13, 0, -13 an = a1 + (n-1)d a25 = 26 + (16)(-13) =-182 a25 = 26 + (17 -1)(-13) Start with the explicit sequence formula Find the common difference between the values. Plug in known values Simplify Common difference a2 - a1 = 13 – 26 = -13
- 8. Ex 3: Find the first term of an arithmetic sequence if the 9th term is 72 and the common difference is 5. an = a1 + (n-1)d 72= a1 + (9-1)5 72= a1 + (8)5 72= a1 + 40 32= a1
- 9. Ex. 4: Suppose you have saved $75 towards the purchase of a new tablet. You plan to save at least $12 from mowing your neighbor’s yard each week. In all, what is the minimum amount of money you will have in 26 weeks? an = a1 + (n-1)d a26 = 75 + (26)12 =$387 difference = 12 a26 = 75 + (27 -1)12 Start with the explicit sequence formula Find the common difference between the values. You will save $12 a week so this is your difference. Substitute in known values Simplify WAIT: Why 27 and not 26 for n ? The first term in the sequence, 75, came before the weeks started (think of it as week 0). Therefore you want one more week in your formula to account for the $75 that you had before you started saving. V. Application of Arithmetic Sequence
- 10. VI. Arithmetic Means Arithmetic Means: the terms between any two nonconsecutive terms of an arithmetic sequence. Example: 17, 10, 3, -4, -11, -18, … Between 10 and -18 there are three arithmetic means 3, -4, -11.
- 11. Ex 2: Find three arithmetic means between 8 and 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. • 8, __, __, __, 14 • a1 = 8, a5 = 14, & n = 5 • 14 = 8 + d(5 - 1) • 14 = 8 + d(4) • 6 = 4d • 1.5 = d • 8, __, __, __, 14 so to find our means we just add 1.5 starting with 8. • 8, 9.5, 11, 12.5, 14
- 12. VII. Arithmetic Series • A series is the expression for the sum of the terms of a sequence, not just “what is the next term?” Ex: 6, 9, 12, 15, 18 . . . This is a list of the numbers in the pattern an not a sum. It is a sequence. Note it goes on forever, so we say it is an infinite sequence. Ex: 6 + 9 + 12 + 15 + 18 Note: if the numbers go on forever, it is infinite; if it has a definitive ending it is finite. Here we are adding the values. We call this a series. Because it does not go on forever, we say it is a finite series.
- 13. Sum of a Finite Arithmetic Series ) ( 2 1 n n a a n S ) 47 2 ( 2 6 n S Where: Sn is the sum of all the terms n = number of terms a1 = first term an = last term For Example, 2 + 11 + 20 + 29 + 38 + 47 = 147 119 ) 29 5 ( 2 7 n S 147 Let’s try one: evaluate the series: 5, 9, 13,17,21,25,29
- 14. • Ofelia sells houses in a new development. She makes a commission of $3750 on the sale of her first house. For each additional house sold, her commission increases by $500. Thus on her next house she will make $4250. How many houses will she have to sell for her total commission to be at least $65,000? • In this situation, her commission is increasing by the same amount each time, and we are asking for the sum of all her commissions. Therefore, this represents an arithmetic series. We are solving for n. VIII. Application of Arithmetic Series
- 15. • Because we have two unknowns (n and an ), we need to substitute something in for one of them. We can substitute • a1 + (n-1)d for an ) ( 2 1 n n a a n S 1 1 ( ( ( 1) )) 2 n n S a a n d 1 (2 ( 1) ) 2 n n S a n d 65000 (2(3750) ( 1)(500)) 2 n n 65000 (7500 500 500) 2 n n
- 16. 65000 (7000 500 ) 2 n n 2 130000 7000 500 n n 2 0 500 7000 130000 n n 2 0 14 260 n n n = 10.58; n=-24.58 She will need to sell 11 houses to make at lest $65,000 a year.