S.3 Geometric Sequences and Series

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68 section S S. Geometric In the previous section, we studied sequences where each term was obtained by adding a constant number to the previous term. In this section, we will take interest in sequences where each term is obtained by multiplying the previous term by a constant number. Such sequences are called geometric. For example, the sequence, 2,, 8, is geometric because each term is multiplied by 2 to obtain the next term. Equivallently, the ratios between consecutive terms of this sequence are always 2. Definition 2. A sequence {aa nn } is called geometric if the quotient rr = aa nn+ aa nn the sequence is constantly the same. The general term of a geometric sequence is given by the formula aa nn = aa rr nn The quotient rr is referred to as the common ratio of the sequence. of any consecutive terms of Similarly as in the previous section, we can be visualize geometric sequences by plotting their values in a system of coordinates. For instance, Figure presents the graph of the sequence,2,, 8,. The common ratio of 2 causes each cosecutive point of the graph to be plotted twice as high as the previous one, and the slope between the nn-th and (nn + )-st point to be exactly equal to the value of aa nn. Generally, the slope between the nn-th and (nn + )-st point of any geometric sequence is proportional to the value of aa nn. This property characterises exponential functions. Hence, geometric sequences are exponential in nature. To develop the formula for the general term, we observe the pattern so aa = aa aa 22 = aa rr aa = aa rr 22 aa = aa rr aa nn = aa rr nn This is the general term of a geometric sequence aa nn 9 aa 8 7 6 5 aa aa 22 22 aa rr = 2 rr = 2 rr = 2 22 Figure nn Particularly, the general term of the sequence, 2,, 8, is equal to aa nn = 22 nn, because aa = and rr = 2 (the ratios of consecutive terms are constantly equal to 2). Note: To find the common ratio of a geometric sequence, divide any of its terms by the preceeding term. Identifying Geometric Sequences and Writing Their General Terms Determine whether the given sequence {aa nn } is geometric. If it is, then write a formula for the general term of the sequence. a.,,,, b.,,,, 2 8 6 6 9 2

section S 69 a. After calculating ratios of terms by their peceding terms, we notice that they are always equal to 2. Indeed, aa 2 = aa = 2, aa aa 2 = 8 =, and so on. Therefore, the given 2 sequence is geometric with aa = and the common ratio rr =. 2 2 To find its general term, we follow the formula aa nn = aa rr nn. This gives us aa nn = 2 2 nn = ( )nn 22 nn. b. Here, the ratios of terms by their peceding terms, aa 2 aa = the same. So the sequence is not geometric. 6 = and aa = 2 aa 2 9 = 2, are not 6 Finding Terms of a Gometric Sequence Given the information, write out the first five terms of the geometric sequence {aa nn }. Then, find the 8-th term aa 8. a. aa nn = 5( 2) nn b. aa =, rr = 2 a. To find the first five terms of this sequence, we evaluate aa nn for nn =,2,,,5. aa = 5( 2) 00 = 5 aa 22 = 5( 2) = 0 aa = 5( 2) 22 = 20 aa = 5( 2) = 0 aa 55 = 5( 2) = 80 So, the first five terms are 55,, 2222,, and 8888. The 8-th term equals aa 88 = 5( 2) 77 = 666666. b. To find the first five terms of a geometric sequence with aa =, rr = 22, we substitute these values into the general term formula aa nn = aa rr nn = 22 nn, and then evaluate it for nn =,2,,,5. This gives us aa = 2 00 = aa 22 = 2 = 2 aa = 2 22 = aa = 2 = 8 9 aa 55 = 2 = 6 27 Geometric

70 section S So, the first five terms are, 22,, 88, and. 99 2222 The 8-th term equals aa 88 = 2 77 = 66666666. Finding the Number of Terms in a Finite Geometric Sequence Determine the number of terms in the geometric sequence,,9, 27,,729. Since the common ratio rr of this sequence is and the first term aa =, then the nn-th term aa nn = ( ) nn. Since the last term is 729, we can set up the equation which can be written as aa nn = ( ) nn = 729, ( ) nn = ( ) 6 This equation holds if nn = 6, which gives us nn = 7. So, there are 7 terms in the given sequence. Finding Missing Terms of a Geometric Sequence Given the information, determine the values of the indicated terms of a geometric sequence. a. aa = 5 and aa 6 = 5; find aa and aa 8 b. aa = 200 and aa 5 = 50; find aa if aa > 0 a. As before, let rr be the common ratio of the given sequence. Using the general term formula aa nn = aa rr nn for nn = 6 and nn =, we can set up a system of two equations in two variables, rr and aa : 5 = aa rr 5 5 = aa rr 2 To solve this system, let s divide the two equations side by side, obtaining which gives us 27 = rr, rr =. Substituting this value to equation (2), we have 5 = aa ( ) 2, which gives us aa = 55 99.

section S 7 To find value aa 8, we substitute aa = 5, rr =, and nn = 8 to the formula aa 2 nn = aa rr nn. This gives us aa 88 = 5 9 ( )7 =. b. Let rr be the common ratio of the given sequence. Since aa 5 = aa rr and aa = aa rr, then aa 5 = aa rr 2. Hence, rr 2 = aa 5. Therefore, aa rr = ± aa 5 aa = ± 50 200 = ± = ± 2. () (2) Since aa, aa > 0 and aa = aa rr, we choose the positive rr-value. So we have aa = aa rr = 200 =. 2 Remark: A geometric mean of two quantities aa and bb is defined as aaaa. Notice that aa = 00 = 50 200 = aa aa 55, so aa is indeed the geometric mean of aa and aa 55. Generally, for any nn >, we have 2 aa nn = aa nn rr = aa nn rr 2 = aa nn aa nn rr 2 = aa nn aa nn+, so every term (except for the first one) of a geometric sequence is the geometric mean of its adjacent terms. Partial Sums Similarly as with arithmetic sequences, we might be interested in evaluating the sum of the first nn terms of a geometric sequence. To find partial sum SS nn of the first nn terms of a geometric sequence, we line up formulas for SS nn and rrrr nn as shown below and then add the resulting equations side by side. SS nn = aa + aa rr + aa rr 22 + + aa rr nn rrss nn = aa rr aa rr 22 aa rr nn aa rr nn the terms of inside columns add to zero, so they cancel each other out So, we obtain ( rr)ss nn = aa aa rr nn, which gives us as long as rr. SS nn = aa ( rr nn ), rr () Geometric

72 section S Observe that If rr <, then the value of rr nn gets closer and closer to zero for larger and larger n (we write: rr nn 0 for nn ). This means that the sum of all infinitely many terms of such a sequence exists and is equal to SS = aa rr () If rr >, then the value of rr nn grows without bound for larger and larger n. Therefore, the sum SS of all terms of such a sequence does not have a finite value. We say that such a sum does not exist. If rr =, then the sum SS becomes aa + aa + aa +, or aa aa + aa. Neither of these sums has a finite value, unless aa = 0. Hence overall, if rr, then the sum SS of a nonzero geometric sequence does not exist. Finding a Partial Sum of a Geometric Sequence a. Find SS 6, for the geometric sequence with aa = 0.5 and rr = 0.. b. Evaluate the sum + 2 9. a. Using formula () for nn = 6, aa = 0.5 and rr = 0., we calculate SS 66 = 0.5( 0.6 ) 0. = 0.5 0.999999 0.9 = 00. 555555555555. b. First, we observe that the given series is geometric with aa = and rr =. Equating the formula for the general term to the last term of the sum aa rr nn = nn = 9 = 9 and comparing the exponents, nn = 9 allows us to find the number of terms nn =. Now, we are ready to calculate the sum of the given series 0 SS = 00. 555555555555

section S 7 Evaluating Infinite Geometric Series Decide wether or not the overall sum SS of each geometric series exists and if it does, evaluate it. a. 9 + 27 8 + b. 2 8 ii=0 2 ii a. Since the common ratio of this series is rr = 9 2 = = >, then the sum SS 2 2 does not exist. b. This time, rr = 2 <, so the sum SS exists and can be calculated by following the formula (). Using aa = and rr = 2, we have So, ii=0 2 ii = 99. SS = 2 = = 99 Using Geometric in Application Problems When dropped from a certain height, a ball rebounds of the original height. mm a. How high will the ball rebound after the fourth bounce if it was dropped from a height of meters? Round the answer to the nearest centimeter. b. Find a formula for the rebound height of the ball after its nn-th bounce. c. Assuming that the ball bounce forever, what is the total vertical distance traveled by the ball? a. Let h nn represents the ball s rebound height after the nn-th bounce, where nn N. Since the ball rebounds of the previous height, we have h = h 22 = h 22 = h = h 2 = h = h =.99 mm 9999 cccc After the fourth bounce, the ball will rebound approximately 95 centimeters. b. Notice that the formulas developed in solution to Example a follow the pattern Geometric

7 section S h nn = nn So this is the formula for the rebound height of the ball after its nn-th bounce. c. Let h 0 = represents the vertical distance before the first bounce. To find the total verical distance DD traveled by the ball, we add the vertical distance h 0 before the first bounce, and twice the vertical distances h nn after each bounce. So we have DD = h 0 + h nn = + nn Applying the formula aa rr nn= nn= for the infinite sum of a geometric series, we calculate 9 DD = + = + 9 = + 9 = + 9 = mm Thus, the total vertical distance traveled by the ball is 2 meters. S. Exercises Vocabulary Check Fill in each blank with appropriate formula, one of the suggested words, or with the most appropriate term or phrase from the given list: consecutive, geometric mean, ratio.. A geometric sequence is a sequence in which there is a constant between consecutive terms. 2. The sequence, 2, 6, 2, iiii / iiii nnnnnn the same. geometric because the ratio between terms is not. The general term of a geometric sequence is given by the formula aa nn =.. If the absolute value of the common ratio of a geometric sequence is than, then the llllllll / mmmmmmmm sum of the associated geometric series exists and it is equal to SS =. 5. The expression aaaa is called the of aa and bb. Concept Check True or False? 6. The sequence,,,, is a geometric sequence. 9 7. The common ratio for 0.05, 0.0505, 0.050505, is 0.05. 8. The series ( 2 ii ) 7 ii= is a geometric series.

section S 75 9. The nn-th partial sum SS nn of any finite geometric series exists and it can be evaluated by using the formula SS nn = aa ( rrnn ). rr Concept Check Identify whether or not the given sequence is geometric. If it is, write a formula for its n-th term. 0. 0,, 9, 27,., 5, 25,25, 2. 9,,,,., 2,,,.,,,, 5. 0.9, 0.09, 0.009,0.0009, 6. 8, 27, 9,, 7., 2,, 8, 8., 5, 25, 6 25, Concept Check Given the information, write out the first four terms of the geometyric sequence {aa nn }. Then, find the 8-th term aa 8. 9. aa nn = 2 nn 20. aa nn = ( 2) nn 2. aa = 6, rr = 22. aa = 5, rr = 2. aa =, aa 2 = 6 2. aa = 00, aa 2 = 0 Concept Check Find the number of terms in each geometric sequence. 25., 2,,, 02 26. 20, 0, 5,, 5 28 27., 2,,, 2 28.,,,, 2 29. 6, 2, 2,, 2 8 0. 2, 2, 6,, 2 Given the information, find the indicated term of each geometric sequence.. aa 2 = 0, rr = 0.; aa 5 2. aa =, aa = 8; aa 0. 2, 2, 2, 2, ; aa 50., 2,, ; aa 2 5. aa = 6, aa = 2 9 ; aa 8 6. aa = 9, aa 6 = 27; aa 9 7. aa = 2, aa 7 = 2 ; aa if aa > 0 8. aa 5 = 8, aa 8 = 8 ; aa 0 Given the geometric sequence, evaluate the indicated partial sum. Round your answer to three decimal places, if needed. 9. aa nn = 5 2 nn ; SS 6 0. aa nn = 2 nn ; SS 0. 2, 6, 8, ; SS 8 2. 6,, 2, ; SS 2. + 5 + 5 2 + + 5 5. + 2 9 5. 7 ii= 2(.05) ii 6. 0 (2) ii ii= Geometric

76 section S Decide wether or not the infinite sum SS of each geometric series exists and if it does, evaluate it. 7. + + 9 + 27 + 8. 5 + 25 6 25 6 + 9. +.02 +.02 2 +.02 + 50. + 0.8 + 0.8 2 + 0.8 + 5. ii= (0.6) ii 52. 2 ii= 5 (.)ii 5. 2 ii= ii 5. 2 ii= ii Analytic Skills Solve each problem. 55. A company is offering a job with a salary of $0,000 for the first year and a 5% raise each year after that. If that 5% raise continues every year, find the amount of money you would earn in the 0 th year of your career. 56. Suppose you go to work for a company that pays one penny on the first day, 2 cents on the second day, cents on the third day and so on. If the daily wage keeps doubling, what will your wage be on the 0 th day? What will your total income be for working 0 days? 57. Suppose a deposit of $2000 is made at the beginning of each year for 5 years into an account paying 2% compounded annually. What is the amount in the account at the end of the forty-fifth year? 58. A father opened a savings account for his daughter on her first birthday, depositing $000. Each year on her birthday he deposits another $000, making the last deposit on her 2st birthday. If the account pays.% interest compounded annually, how much is in the account at the end of the day on the daughter s 2st birthday? 59. A ball is dropped from a height of 2 m and bounces 90% of its original height on each bounce. a. How high off the floor is the ball at the top of the eighth bounce? b. Asuming that the ball moves only vertically, how far has it traveled when it hits the ground for the eighth time? 60. Suppose that a ball always rebounds 2 of the distance from which it falls. If this ball is dropped from a height of 9 ft, then approximately how far does it travel before coming to rest? Assime that the ball moves only vertically. 6. Suppose the midpoints of a unit square ss (with the length of each side equal to one) are connected to form another square, ss 2, as in the accompanying figure. Suppose we continue indefinitely the process of creating a new square, ss nn+, by connecting the midpoints of the previous square, ss nn. Calculate the sum of the areas of the infinite sequence of squares {ss nn }.