HOME

David Terr
Ph.D. Math, UC Berkeley

 

Home >> Pre-Calculus >> 10. Sequences, Induction, and Probability

<< 10.2. Arithmetic Sequences

>> 10.4. Mathematical Induction

 

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) geometric  series formula

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:

geometric series with |r|<1

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) infinite geometric series

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.

 

Home >> Pre-Calculus >> 10. Sequences, Induction, and Probability

<< 10.2. Arithmetic Sequences

>> 10.4. Mathematical Induction