HOME David Terr Ph.D. Math, UC Berkeley 10.3. Geometric Sequences Another important type of sequence are geometric sequences. A geometric sequence is a sequence of the form a0 = a, a1 = ar, a2 = ar2, a3 = ar3, ..., where a and r are nonzero constants. We see that the term with index k is ak = ark. Thus, like with arithmetic sequences, it is easy to compute terms of a geometric sequence with known index if we know a and r. But a and r are easy to compute. Assuming the sequence begins with index 0, a is the first term of the sequence. Also, r is the ratio of consecutive terms in the sequence (r = ak+1 / ak).   Example 1: For each of the following geometric sequences, compute a5 and a10. (Assume each sequence starts with index 0.) (a) 1, 2, 4, 8, ... (b) 1, 10, 100, 1000, ... (c) 1, 1/2, 1/4, 1/8, ... (d) 1, -1/2, 1/4, -1/8, ... (e) 3, 9, 27, 81, ... (f) 3, 6, 12, 24, ... Solution: (a) For this sequence, we have a = 1 and r = 2. Thus, we have ak = 2k, implying a5 = 25 = 32 and a10 = 210 = 1024. (b) For this sequence, we have a = 1 and r = 10. Thus, we have ak = 10k, implying a5 = 105 = 100,000 and a10 = 1010 = 10,000,000,000. (c) For this sequence, we have a = 1 and r = 1/2. Thus, we have ak = (1/2)k = 1/2k, implying a5 = 1/25= 1/32 and a10 = 1/210 = 1 / 1024. (d) For this sequence, we have a = 1 and r = -1/2. Thus, we have ak = (-1/2)k = (-1)k / 2k, implying a5 = (-1)5/ 25= -25 = -1/32 and a10 = (-1)10 / 210 = 1 / 1024. (e) For this sequence, we have a = r = 3. Thus, we have ak = 3 * 3k = 3k+1, implying a5 = 36 = 729 and a10 = 311 = 177,147. (f) For this sequence, we have a = 3 and r = 2. Thus, we have ak = 3 * 2k, implying a5 = 3 * 25 = 96 and a10 = 3 * 210 = 3072.   As is the case with arithmetic sequences, there is a formula for computing series of geometric sequences, known as geometric series. The formula is as follows. Here we assume a and r are arbitrary real numbers with r≠1. (10.3.1) It is fairly easy to derive Equation (10.3.1). Consider the geometric series above; call it S. Writing out all the terms, we have S = a + ar + ar2 + ar3 + ... + arn. Multiplying this series by r, we find rS = ar + ar2 + ar3 + ... + arn+1. Now subtracting the first equation from the second, we find (r - 1)S = ar + ar2 + ar3 + ... + arn+1 - a - ar - ar2 + ar3 - ... - arn = arn+1 - a = a(rn+1 - 1). Note that all the middle terms cancel, leaving behind only the last term of rS minus the first term of S. Dividing both sides by r - 1 leads to (10.3.1).   Example 2: Use Equation (10.3.1) to compute the sum of the terms of each sequence in Example 1 with index up to 10, i.e. the first 11 terms of each sequence. Solution: (a) We have S = (1)(211 - 1) / (2 - 1) = 211 - 1 = 2048 - 1 = 2047. (b) We have S = (1)(1011 - 1) / (10 - 1) = (1011 - 1) / 9 = 99,999,999,999 / 9 = 11,111,111,111. (c) We have S = (1)[(1/2)11 - 1] / (1/2 - 1) = (1 - 1/211) / (1/2) = (211 - 1) / 210 = 2047 / 1024. (d) We have S = (1)[(-1/2)11 - 1] / (-1/2 - 1) = (1 + 1/211) / (3/2) = (211+ 1) / (3 * 210) = 2049 / 3072 = 683 / 1024. (e) We have S = (3)(311 - 1) / (3 - 1) = (3)(311 - 1) / 2= (3)(177,146) / 2 = 265,719. (f) We have S = (3)(211 - 1) / (2 - 1) = (3)(211 - 1) = (3)(2047) = 6141.   It is instructive to consider the case of geometric series in which |r| < 1. For these sequences, it is better to write Equation (10.3.1) as follows: Note that as n gets large, rn+1 gets smaller and smaller, approaching zero as n goes to infinity. Thus, in the limit in which n goes to infinity, the series can be expressed as follows: (10.3.2) Thus we see that if |a| is less than 1, the infinite geometric series has a finite value. Such geometric series are said to converge.   Example 3: Use Equation (10.3.2) to compute the following infinite series: (a) S = 1 + 1/2 + 1/4 + 1/8 + ... (b) S = 3/10 + 3/100 + 3/1000 + ... (c) S = 1 - 1/2 + 1/4 - 1/8 +- ... (d) S = 2/3 + 4/9 + 8/27 + ... Solution: (a) For this series, we have a = 1 and r = 1/2, whence S = 1/(1 - 1/2) = 1/(1/2) = 2. (b) For this series, we have a = 3/10 and r = 1/10, whence S = (3/10) / (1 - 1/10) = (3/10) / (9/10) = 3/9 = 1/3. (c) For this series, we have a = 1 and r = -1/2, whence S = 1/(1 + 1/2) = 1/(3/2) = 2/3. (c) For this series, we have a = r = 2/3, whence S = (2/3) / (1 - 2/3) = (2/3) / (1/3) = 2. Copyright © 2007-2009 - MathAmazement.com. All Rights Reserved.