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7.7.1: Finding the nth Term Given the Common Ratio and the First Term

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    14803
  • Geometric Sequences and Finding the nth Term Given the Common Ratio and the First Term

    The following sequence shows the distance (in centimeters) a pendulum travels with each successive swing. Write a general rule for the geometric sequence.

    80, 72, 64.8, 58.32, ...


    Geometric Sequence

    geometric sequence is a sequence in which the ratio between any two consecutive terms, \(\ \frac{a_{n}}{a_{n-1}}\), is constant. This constant value is called the common ratio. Another way to think of this is that each term is multiplied by the same value, the common ratio, to get the next term.

    Let's consider the sequence 2, 6, 18 ,54, …

    Is this sequence geometric? If so, what is the common difference?

    If we look at each pair of successive terms and evaluate the ratios, we get \(\ \frac{6}{2}=\frac{18}{6}=\frac{54}{18}=3\) which indicates that the sequence is geometric and that the common ratio is 3.

     

    Now let’s see if we can develop a general rule ( \(\ n^{t h}\) term) for this sequence. Since we know that each term is multiplied by 3 to get the next term, let’s rewrite each term as a product and see if there is a pattern.

    \(\ \begin{array}{l}
    a_{1}=2 \\
    a_{2}=a_{1}(3)=2(3)=2(3)^{1} \\
    a_{3}=a_{2}(3)=2(3)(3)=2(3)^{2} \\
    a_{4}=a_{3}(3)=2(3)(3)(3)=2(3)^{3}
    \end{array}\)

    This illustrates that the general rule is \(\ a_{n}=a_{1}(r)^{n-1}\), where \(\ r\) is the common ratio. This even works for the first term since \(\ a_{1}=2(3)^{0}=2(1)=2\). 

    Now, let's write a general rule for the geometric sequence 64, 32, 16, 8, …

    From the general rule above we can see that we need to know two things: the first term and the common ratio to write the general rule. The first term is 64 and we can find the common ratio by dividing a pair of successive terms, \(\ \frac{32}{64}=\frac{1}{2}\). The \(\ n^{t h}\) term rule is thus \(\ a_{n}=64\left(\frac{1}{2}\right)^{n-1}\).

    Finally, let's find the \(\ n^{t h}\) term rule for the sequence 81, 54, 36, 24, … and hence find the \(\ 12^{t h}\) term.

    The first term here is \(\ 81\) and the common ratio, \(\ r\), is \(\ \frac{54}{81}=\frac{2}{3}\). The \(\ n^{t h}\) term rule is \(\ a_{n}=81\left(\frac{2}{3}\right)^{n-1}\). Now we can find the \(\ 12^{t h}\) term \(\ a_{12}=81\left(\frac{2}{3}\right)^{12-1}=81\left(\frac{2}{3}\right)^{11}=\frac{2048}{2187}\).

    Use the graphing calculator for the last step and MATH > Frac your answer to get the fraction. We could also use the calculator and the general rule to generate terms seq(81(2/3)(x−1),x,12,12). Reminder: the seq( ) function can be found in the LIST (2nd STAT) Menu under OPS. Be careful to make sure that the entire exponent is enclosed in parenthesis.


    Examples

    Example 1

    Earlier, you were asked to write a general rule for the sequence 80, 72, 64.8, 58.32, ...

    Solution

    We need to know two things, the first term and the common ratio, to write the general rule. The first term is 80 and we can find the common ratio by dividing a pair of successive terms, \(\ \frac{72}{80}=\frac{9}{10}\). The \(\ n^{t h}\) term rule is thus \(\ a_{n}=80\left(\frac{9}{10}\right)^{n-1}\)

    For Examples 2-4, identify which of the sequences are geometric sequences. If the sequence is geometric, find the common ratio.

    Example 2

    5, 10, 15, 20, …

    Solution

    arithmetic

    Example 3

    1, 2, 4, 8, …

    Solution

    geometric, \(\ r=2\)

    Example 4

    243, 49, 7, 1, …

    Solution

    geometric, \(\ r=\frac{1}{7}\)

    Example 5

    Find the general rule and the \(\ 20^{t h}\) term for the sequence 3, 6, 12, 24, …

    Solution

    The first term is 3 and the common ratio is \(\ r=\frac{6}{3}=2\) so \(\ a_{n}=3(2)^{n-1}\).

    The \(\ 20^{t h}\) term is \(\ a_{20}=3(2)^{19}=1,572,864\).

    Example 6

    Find the \(\ n^{t h}\) term rule and list terms 5 thru 11 using your calculator for the sequence −1024, 768, −432, −324, …

    Solution

    The first term is -1024 and the common ratio is \(\ r=\frac{768}{-1024}=-\frac{3}{4}\) so \(\ a_{n}=-1024\left(-\frac{3}{4}\right)^{n-1}\). 

    Using the calculator sequence function to find the terms and MATH > Frac,

    \(\ \text { seq }\left(-1024(-3 / 4)^{\wedge}(x-1), x, 5,11\right)=\left\{\begin{array}{l}
    -324 & 243 & -\frac{729}{4} & \frac{2187}{16} & -\frac{6561}{256} & \frac{19683}{256} & \left.-\frac{59049}{1024}\right\}
    \end{array}\right.\)

    Example 7

    Find the value of a 10 year old car if the purchase price was $22,000 and it depreciates at a rate of 9% per year.

    Solution

    The first term (value of the car after 0 years) is $22,000. The common ratio is 1−.09 or 0.91.

    The value of the car after \(\ n\) years can be determined by \(\ a_{n}=22,000(0.91)^{n}\). For 10 years we get \(\ a_{10}=22,000(0.91)^{10}=8567.154599 \approx \$ 8567\)


    Review

    Identify which of the following sequences are arithmetic, geometric or neither.

    1. \(\ 2,4,6,8, \ldots\)
    2. \(\ \frac{1}{2}, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}, \ldots\)
    3. \(\ 1,2,4,7, \ldots\)
    4. \(\ 24,-16, \frac{32}{3},-\frac{64}{9}, \ldots\)
    5. \(\ 10,5,0,-5, \ldots\)
    6. \(\ 3,4,7,11, \ldots\)

    Given the first term and common ratio, write the \(\ n^{t h}\) term rule and use the calculator to generate the first five terms in each sequence.

    1. \(\ a_{1}=32\) and \(\ r=\frac{3}{2}\)
    2. \(\ a_{1}=-81\) and \(\ r=-\frac{1}{3}\)
    3. \(\ a_{1}=7\) and \(\ r=2\)
    4. \(\ a_{1}=\frac{8}{125}\) and \(\ r=-\frac{5}{2}\)

    Find the \(\ n^{t h}\) term rule for each of the following geometric sequences.

    1. \(\ 162,108,72, \ldots\)
    2. \(\ -625,-375,-225, \ldots\)
    3. \(\ \frac{9}{4},-\frac{3}{2}, 1, \ldots\) 
    4. \(\ 3,15,75, \ldots\)
    5. \(\ 5,10,20, \ldots\)
    6. \(\ \frac{1}{2},-2,8, \ldots\)

    Use a geometric sequence to solve the following word problems.

    1. Rebecca inherited some land worth $50,000 that has increased in value by an average of 5% per year for the last 5 years. If this rate of appreciation continues, about how much will the land be worth in another 10 years?
    2. A farmer buys a new tractor for $75,000. If the tractor depreciates in value by about 6% per year, how much will it be worth after 15 years?

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 11.8.

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