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College Algebra & Trigonometry 4th EDITION and Precalculus 10TH EDITION 7.2 - 1 11.2 Arithmetic Sequences and Series Arithmetic Sequences Arithmetic Series 7.2 - 2 Arithmetic Sequences A sequence in which each term after the first is obtained by adding a fixed number to the previous term is an arithmetic sequence (or arithmetic progression). The fixed number that is added is the common difference. The sequence 5,9,13,17,21, is an arithmetic sequence since each term after the first is obtained by adding 4 to the previous term. 7.2 - 3 Arithmetic Sequences That is, 954 13 9 4 17 13 4 21 17 4, and so on. The common difference is 4. 7.2 - 4 Arithmetic Sequences If the common difference of an arithmetic sequence is d, then by the definition of an arithmetic sequence, d an 1 an , Common difference d for every positive integer n in the domain of the sequence. 7.2 - 5 Example 1 FINDING THE COMMON DIFFERENCE Find the common difference, d, for the arithmetic sequence 9, 7, 5, 3, 1, Solution We find d by choosing any two adjacent terms and subtracting the first from the second. Choosing – 7 and – 5 gives Be careful when d 5 ( 7) 2. subtracting a negative number. Choosing – 9 and – 7 would give the same result. d 7 ( 9) 2. 7.2 - 6 Example 2 FINDING TERMS GIVEN a1 AND d Find the first five terms for each arithmetic sequence. a. The first term is 7, and the common difference is – 3. Solution a1 7 Start with a1 = 7. a2 7 ( 3) 4 Add d = – 3. a3 4 ( 3) 1 Add – 3. a4 1 ( 3) 2 Add – 3. a5 2 ( 3) 5 Add – 3. 7.2 - 7 Example 2 FINDING TERMS GIVEN a1 AND d Find the first five terms for each arithmetic sequence. b. a1 = – 12, d = 5 Solution a1 12 a2 12 5 7 Start with a1. Add d = 5. a3 7 5 2 a4 2 5 3 a5 3 5 8 7.2 - 8 Look At It This Way If a1 is the first term of an arithmetic sequence and d is the common difference, then the terms of the sequence are given by a1 a1 a2 a1 d a3 a2 d a1 d d a1 2d a4 a3 d a1 2d d a1 3d a5 a1 4d a6 a1 5d, and, by this pattern, an a1 (n 1)d. 7.2 - 9 nth Term of an Arithmetic Sequence In an arithmetic sequence with first term a1 and common difference d, the nth term, is given by an a1 (n 1)d. 7.2 - 10 Example 3 FINDING TERMS OF AN ARITHMETIC SEQUENCE Find a13 and an for the arithmetic sequence – 3, 1, 5, 9, … Solution Here a1 = – 3 and d = 1 – (– 3) = 4. To find a13 substitute 13 for n in the formula for the nth term. a1 a1 (n 1)d Work inside parentheses first. a13 a1 (13 1)d n = 13 a13 3 (12)4 Let a1 = – 3, d = 4. a13 3 48 Simplify. a13 45 7.2 - 11 Example 3 FINDING TERMS OF AN ARITHMETIC SEQUENCE Find a13 and an for the arithmetic sequence – 3, 1, 5, 9, … Solution Find an by substituting values for a1 and d in the formula for an. an 3 (n 1) 4 Let a1 = – 3, d = 4. an 3 4n 4 Distributive property an 4n 7 Simplify. 7.2 - 12 Example 4 FINDING TERMS OF AN ARTHEMTIC SEQUENCE Find a18 and an for the arithmetic sequence having a2 = 9 and a3 = 15. Solution Find d first; d = a3 – a2 = 15 – 9 = 6. Since a2 a1 d , 9 a1 6 Let a2 = 9, d = 6. a1 3. 7.2 - 13 Example 4 FINDING TERMS OF ARTHEMTIC SEQUENCE Find a18 and an for the arithmetic sequence having a2 = 9 and a3 = 15. Solution d = a3 – a2 = 15 – 9 = 6. Then, a18 3 (18 1)6 a18 105, Formula for an; a1 = 3, n = 18, d = 6. 7.2 - 14 Example 4 FINDING TERMS OF ARTHEMTIC SEQUENCE Find a18 and an for the arithmetic sequence having a2 = 9 and a3 = 15. Solution d = a3 – a2 = 15 – 9 = 6. and an 3 (n 1)6 an 3 6n 6 Distributive property. an 6n 3. 7.2 - 15 Example 5 FINDING THE FIRST TERM OF AN ARITHMETIC SEQUENCE Suppose that an arithmetic sequence has a8 = – 16 and a16 = – 40. Find a1 Solution Since a16 = a8 + 8d, it follows that 8d a16 a8 40 ( 16) 24, And so d = – 3. To find a1, use the equation a8 = a1 + 7d. 16 a1 7d Let a8 = – 16. 16 a1 7( 3) Let d = – 3. a1 5 7.2 - 16 FINDING THE FIRST TERM OF AN ARITHMETIC SEQUENCE The graph of any sequence is a scatter diagram. To determine the characteristics of the graph of an arithmetic sequence, start by rewriting the formula for the nth term. an a1 (n 1)d Formula for the nth term. a1 nd d Distributive property. dn (a1 d ) Commutative & associative properties. dn c Let c = a1 – d. 7.2 - 17 FINDING THE FIRST TERM OF AN ARITHMETIC SEQUENCE The points in the graph of an arithmetic sequence are determined by (n) = dn + c, where n is a natural number. Thus, the points in the graph of must lie on the line y dx c. Slope y-intercept 7.2 - 18 FINDING THE FIRST TERM OF AN ARITHMETIC SEQUENCE For example, the sequence shown here is an arithmetic sequence because the points that comprise its graph are collinear (lie on a line). The slope determined by these points is 2, so the common difference d equals 2. 7.2 - 19 FINDING THE FIRST TERM OF AN ARITHMETIC SEQUENCE On the other hand, the sequence bn shown here is not an arithmetic sequence because the points are not collinear. 7.2 - 20 Example 6 FINDING THE nth TERM FROM A GRAPH Find a formula for the nth term of the sequence an shown here. What are the domain and range of this sequence? Solution The points in this graph lie on a line, so the sequence is arithmetic. The equation of the dashed line shown here is y = – .5x + 4, so the nth term of this sequence is determined by an .5n 4. 7.2 - 21 Example 6 FINDING THE nth TERM FROM A GRAPH Find a formula for the nth term of the sequence an shown here. What are the domain and range of this sequence? Solution The sequence is comprised of the points (1,3.5),(2,3),(3,2.5), (4,2),(5,1.5),(6,1). 7.2 - 22 Example 6 FINDING THE nth TERM FROM A GRAPH Find a formula for the nth term of the sequence an shown here. What are the domain and range of this sequence? Solution Thus, the domain of the sequence is given by {1, 2, 3, 4, 5, 6}, and the range is given by {3.5, 3, 2.5, 2, 1.5, 1}. 7.2 - 23 Arithmetic Series The sum of the terms of an arithmetic sequence is an arithmetic series. To illustrate, suppose that a person borrows $3000 and agrees to pay $100 per month plus interest of 1% per month on the unpaid balance until the loan is paid off. The first month, $100 is paid to reduce the loan, plus interest of (.01)3000 = 30 dollars. The second month, another $100 is paid toward the loan, and (.01)2900 dollars is paid for interest 7.2 - 24 Arithmetic Series Since the loan is reduced by $100 each month, interest payments decrease by (.01)100 = 1 dollar each month, forming the arithmetic sequence 30, 29, 28, …, 3, 2, 1. 7.2 - 25 Arithmetic Series The total amount of interest paid is given by the sum of the terms of this sequence. Now we develop a formula to find this sum without adding all 30 numbers directly. Since the sequence is arithmetic, we can write the sum of the first n terms as Sn a1 a1 d a1 2d a1 (n 1)d . 7.2 - 26 Arithmetic Series We used the formula for the general term in the last expression. Now we write the same sum in reverse order, beginning with an and subtracting d. Sn an an d an 2d an (n 1)d 7.2 - 27 Arithmetic Series Adding respective sides of these two equations term by term, we obtain Sn Sn (a1 an ) (a1 an ) or (a1 an ), 2Sn n(a1 an ), since there are n terms of a1 + an on the right. Now solve for Sn to get n Sn (a1 an ). 2 7.2 - 28 Arithmetic Series Using the formula an = a1 + (n – 1)d, we can also write this result for Sn as or n Sn [a1 a1 (n 1)d ], 2 n Sn [2a1 (n 1)d ], 2 which is an alternative formula for the sum of the first n terms of an arithmetic sequence. 7.2 - 29 Sum of the First n Terms of an Arithmetic Sequence If an arithmetic sequence has first term and common difference d, then the sum of the first n terms is given by n n Sn (a1 an ) or Sn [2a1 (n 1)d ]. 2 2 7.2 - 30 The first formula is used when the first and last terms are known; otherwise the second formula is used. For example, in the sequence of interest payments discussed earlier, n = 30, a1 = 30, and an = 1. Choosing the first formula, n Sn (a1 an ), 2 30 gives S30 (30 1) 15(31) 465, 2 so a total of $465 interest will be paid over the 30 months. 7.2 - 31 Example 7 USING THE SUM FORMULAS a. Evaluate S12 for the arithmetic sequence – 9, – 5, – 1, 3, 7, …. Solution We want the sum of the first 12 terms. Using a1 = – 9, n = 12, and d = 4 in the second formula, n Sn [2a1 (n 1)d ], 2 12 gives S12 [2( 9) 11(4)] 156. 2 7.2 - 32 Example 7 USING THE SUM FORMULAS b. Use a formula for Sn to evaluate the sum of the first 60 positive integers. Solution The first 60 positive integers form the arithmetic sequence 1, 2, 3, 4, …, 60. Thus, n = 60, a1 = 1, and a60 = 60, so we use the first formula in the preceding box to find the sum. n Sn (a1 an ) 2 60 S60 (1 60) 1830 2 7.2 - 33 Example 8 USING THE SUM FORMULA The sum of the first 17 terms of an arithmetic sequence is 187. If a17 = – 13, find a1 and d. Solution 17 S17 (a1 a17 ) 2 17 187 (a1 13) 2 22 a1 13 a1 35 Use the first formula for Sn, with n = 17. Let S17 = 187, a17 = – 13. 2 . Multiply by 17 Add 13; rewrite. 7.2 - 34 Example 8 USING THE SUM FORMULA The sum of the first 17 terms of an arithmetic sequence is 187. If a17 = – 13, find a1 and d. Solution Since a17 = a1 + (17 – 1)d, 13 35 16d Let a17 = – 13, a1 = 35. 48 16d Subtract 35. d 3. Divide by 16; rewrite. 7.2 - 35 n Any sum of the form (d i c ), where d and c i 1 are real numbers, represents the sum of the terms of an arithmetic sequence having first term a1 = d(1) + c = d + c and common difference d. These sums can be evaluated using the formulas in this section. 7.2 - 36 Example 9 USING SUMMATION NOTATION Evaluate each sum. 10 a. (4i 8) i 1 Solution This sum contains the first 10 terms of the arithmetic sequence having a1 4 1 8 12, and a10 4 10 8 48. First term. Last term. 10 10 Thus, (4i 8) S10 (12 48) 5(60) 300. 2 i 1 7.2 - 37 Example 9 USING SUMMATION NOTATION Evaluate each sum. 9 b. (4 3k ) k 3 Solution The first few terms are [4 3(3)] [4 3(4)] [4 3(5)] 5 ( 8) ( 11) Thus, a1 = – 5 and d = – 3. If the sequence started with k = 1, there would be nine terms. Since it starts at 3, two of those terms are missing, so there are seven terms and n = 7. Use the second formula for Sn. 9 7 (4 3k ) [2( 5) 6( 3)] 98 2 k 3 7.2 - 38