Exercise 1 Sequences and Series

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1 Section 5 Arithmetic Progressions Does the series converge or diverge? Solution: Again we apply the three tests. 1. We see that the terms are getting smaller, so this test is not conclusive.. The ratio of the terms appears to approach 1, so this test also is not conclusive. 3. The partial sum S n does not seem to approach a limit, but appears to keep growing even after 495 terms. It thus appears that this series diverges. Exercise 1 Sequences and Series Write the first five terms of each series, given the general term. 1. u n 3n. u n n 3 3. u n n 1 n 4. u n n n Deduce the general term of each series. Use it to predict the next two terms Deduce a recursion relation for each series. Use it to predict the next two terms Arithmetic Progressions Recursion Formula We stated earlier that an arithmetic progression (or AP) is a sequence of terms in which each term after the first equals the sum of the preceding term and a constant, called the common difference, d. If a n is any term of an AP, the recursion formula for an AP is as follows: AP: Recursion Formula a n a n1 d 36 Each term of an AP after the first equals the sum of the preceding term and the common difference.

2 704 Chapter 5 Sequences, Series, and the Binomial Theorem Example 10: The following sequences are arithmetic progressions. The common difference for each is given in parentheses: (a) 1, 5, 9, 13,... (d 4) (b) 0, 30, 40, 50,... (d 10) (c) 75, 70, 65, 60,... (d 5) We see that each series is increasing when d is positive and decreasing when d is negative. General Term For an AP whose first term is a and whose common difference is d, the terms are a, a d, a d, a 3d, a 4d,... We see that each term is the sum of the first term and a multiple of d, where the coefficient of d is one less than the number n of the term. So the nth term a n is given by the following equation: The general term is sometimes called the last term, but this is not an accurate name since the AP continues indefi nitely. AP: General Term a n a (n 1)d 37 The nth term of an AP is found by adding the first term and (n 1) times the common difference. Example 11: Find the twentieth term of an AP that has a first term of 5 and common difference of 4. Solution: By Eq. 37, with a 5, n 0, and d 4, a (4) 81 Of course, Eq. 37 can be used to find any of the four quantities (a, n, d, or a n ) given the other three. Example 1: Write the AP whose eighth term is 19 and whose fifteenth term is 33. Solution: Applying Eq. 37 twice gives 33 a 14d 19 a 7d We now have two equations in two unknowns, which we solve simultaneously. Subtracting the second from the first gives 14 7d, so d. Also, a 33 14d 5, so the general term of our AP is and the AP is then a n 5 (n 1) 5, 7, 9,... AP: Sum of n Terms Let us derive a formula for the sum s n of the first n terms of an AP. Adding term by term gives s n a (a d) (a d ) (a n d ) a n (1)

3 Section 5 Arithmetic Progressions 705 or, written in reverse order, s n a n (a n d) (a n d) (a d) a () Adding Eqs. (1) and () term by term gives s n (a a n ) (a a n ) (a a n ) n(a a n ) Dividing both sides by gives the following formula: AP: Sum of n Terms s n n (a a n) 38 The sum of n terms of an AP is half the product of n and the sum of the first and nth terms. Example 13: Find the sum of 10 terms of the AP, 5, 8, 11,... Solution: From Eq. 37, with a, d 3, and n 10, Then, from Eq. 38, a 10 9(3) 9 10 ( 9) s We get another form of Eq. 38 by substituting the expression for a n from Eq. 37, as follows: AP: Sum of n Terms s n n [a (n 1)d] 39 This form is useful for finding the sum without first computing the nth term. Example 14: We repeat Example 13 without first having to find the tenth term. From Eq. 39, 10 [() 9 (3)] s Sometimes the sum is given and we must find one of the other quantities in the AP.

4 706 Chapter 5 Sequences, Series, and the Binomial Theorem Example 15: How many terms of the AP 5, 9, 13,... give a sum of 75? Solution: We seek n so that s n 75. From Eq. 39, with a 5 and d 4, 75 n [(5) (n 1)4] 5n n (n 1) Removing parentheses and collecting terms gives the quadratic equation n 3n 75 0 From the quadratic formula, n ()(75) 11 or 1.5 () We discard the negative root and get 11 terms as our answer. Arithmetic Means The first term a and the last (nth) term a n of an AP are sometimes called the extremes, while the intermediate terms, a, a 3,..., a n 1, are called arithmetic means. We now show, by example, how to insert any number of arithmetic means between two extremes. Example 16: Insert five arithmetic means between 3 and 9. Solution: Our AP will have seven terms, with a first term of 3 and a seventh term of 9. From Eq. 37, 9 3 6d from which d. The progression is then 3, 1, 1, 3, 5, 7, 9 and the five arithmetic means are 1, 1, 3, 5, and 7. Average Value Let us insert a single arithmetic mean between two numbers. Example 17: Find a single arithmetic mean m between the extremes a and b. Solution: The sequence is a, m, b. The common difference d is m a b m. Solving for m gives m b a Dividing by gives m a b which agrees with the common idea of an average of two numbers as the sum of those numbers divided by. Harmonic Progressions A sequence is called a harmonic progression if the reciprocals of its terms form an arithmetic progression. Example 18: The sequence 1 1, 3, 1 5, 1 7, 1 9, 1 11,... is a harmonic progression because the reciprocals of the terms, 1, 3, 5, 7, 9, 11,..., form an AP.

5 Section 5 Arithmetic Progressions 707 It is not possible to derive an equation for the nth term or for the sum of a harmonic progression. However, we can solve problems involving harmonic progressions by taking the reciprocals of the terms and using the formulas for the AP. Example 19: Find the tenth term of the harmonic progression, 3, 5,... Solution: We write the reciprocals of the terms, 1, 3, 5,... and note that they form an AP with a 1 and d 1. The tenth term of the AP is then a n a (n 1)d a 10 1 (10 1)(1) 19 The tenth term of the harmonic progression is the reciprocal of 19, or 19. Harmonic Means To insert harmonic means between two terms of a harmonic progression, we simply take the reciprocals of the given terms, insert arithmetic means between those terms, and take reciprocals again. Example 0: Insert three harmonic means between and. 9 Solution: Taking reciprocals, our AP is 9,,,, 1 In this AP, a 9, n 5, and a 1. We can find the common difference d from the equation 5 a n a (n 1)d 1 9 4d 4 4d d 1 Having the common difference, we can fill in the missing terms of the AP, 9, 7, 5, 3, 1 Taking reciprocals again, our harmonic progression is 9, 7, 5, 3,

6 708 Chapter 5 Sequences, Series, and the Binomial Theorem Exercise Arithmetic Progressions General Term 1. Find the fifteenth term of an AP with first term 4 and common difference 3.. Find the tenth term of an AP with first term 8 and common difference. 3. Find the twelfth term of an AP with first term 1 and common difference Find the ninth term of an AP with first term 5 and common difference. 5. Find the eleventh term of the AP 6. Find the eighth term of the AP 7. Find the ninth term of the AP 9, 13, 17,... 5, 8, 11,... x, x 3y, x 6y, Find the fourteenth term of the AP 1, 6 7, 5 7,... For problems 9 through 1, write the first five terms of each AP. 9. First term is 3 and thirteenth term is First term is 5 and tenth term is Seventh term is 41 and fifteenth term is Fifth term is 7 and twelfth term is Find the first term of an AP whose common difference is 3 and whose seventh term is Find the first term of an AP whose common difference is 6 and whose tenth term is 77. Sum of an Arithmetic Progression 15. Find the sum of the first 1 terms of the AP 3, 6, 9, 1, Find the sum of the first five terms of the AP 1, 5, 9, 13, Find the sum of the first nine terms of the AP 5, 10, 15, 0, Find the sum of the first 0 terms of the AP 1, 3, 5, 7, How many terms of the AP 4, 7, 10,... will give a sum of 375? 0. How many terms of the AP, 9, 16,... will give a sum of 70? Arithmetic Means 1. Insert two arithmetic means between 5 and 0.. Insert five arithmetic means between 7 and Insert four arithmetic means between 6 and Insert three arithmetic means between 0 and 56

7 Section 5 3 Geometric Progressions 709 Harmonic Progressions 5. Find the fourth term of the harmonic progression 3 5, 3 8, 3 11, Find the fifth term of the harmonic progression 4 19, 4 15, 4 11,... Harmonic Means 7. Insert two harmonic means between 7 and Insert three harmonic means between 6 and Applications 9. Loan Repayment: A person agrees to repay a loan of $10,000 with an annual payment of $1,000 plus 8% of the unpaid balance. (a) Show that the interest payments alone form the AP: $800, $70, $640,.... (b) Find the total amount of interest paid. 30. Simple Interest: A person deposits $50 in a bank on the first day of each month, at the same time withdrawing all interest earned on the money already in the account. (a) If the rate is 1% per month, computed monthly, write an AP whose terms are the amounts withdrawn each month. (b) How much interest will have been earned in the 36 months following the first deposit? 31. Straight-Line Depreciation: A certain milling machine has an initial value of $150,000 and a scrap value of $10,000 twenty years later. Assuming that the machine depreciates the same amount each year, find its value after 8 years. 3. Salary or Price Increase: A person is hired at a salary of $40,000 and receives a raise of $,500 at the end of each year. Find the total amount earned during 10 years. 33. Freely Falling Body: A freely falling body falls g metres during the first second, 3g m during the next second, 5g m during the third second, and so on, where g ms. Find the total distance the body falls during the first 10 s. 34. Using the information of problem 33, show that the total distance s fallen in t seconds is s 1 gt. To fi nd the amount of depreciation for each year, divide the total depreciation (initial value scrap value) by the number of years of depreciation. 5 3 Geometric Progressions Recursion Formula A geometric sequence or geometric progression (GP) is one in which each term after the first is formed by multiplying the preceding term by a factor r, called the common ratio. Thus if a n is any term of a GP, the recursion relation is as follows: GP: Recursion Formula a n ra n1 40 Each term of a GP after the first equals the product of the preceding term and the common ratio.