Precalculus by Richard Wright

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Wealth is worthless in the day of wrath, but righteousness delivers from death. Proverbs‬ ‭11‬:‭4‬ ‭NIV‬‬‬

10-04 Geometric Sequences and Series

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.5.5, PC.7.2

tree diagram
Figure 1: Doubling diagram. credit (Wikimedia/Graouilly54)

A Christian wants to spread her love for Jesus to others. She tells three people about Jesus who then they each tell two other people each. If each person tells three people about Jesus who then tell three more people. How many total people have been told about Jesus after the 10th set of people have been told about Him? This type of problem is the sum of a geometric series. But first what is a geometric sequence.

Geometric Sequences

A geometric sequence with a pattern of a common ratio, r. The explicit formula can by derived by looking at a simple example.

$$ 1, 3, 9, 27, 81, 243, … $$

Write the numbers using the pattern which in this example is multiplying by 3.

$$ 1, 1·3, 1·3·3, 1·3·3·3, 1·3·3·3·3, … $$

$$ 1, 1·3^\color{blue}{1}, 1·3^\color{blue}{2}, 1·3^\color{blue}{3}, 1·3^\color{blue}{4}, … $$

The blue terms are 1 less than the term number, so the rule becomes

$$ a_n = 1·3^{n - 1} $$

Notice that the first term was 1 and the common ratio was 3, replace those numbers in the rule to get the general formula for the nth term.

$$ a_n = a_1 · r^{n - 1} $$

Explicit Rule for Geometric Sequences

$$ a_n = a_1 · r^{n - 1} $$

where a1 is the first term and r is the common ratio

Recursive Rule for Geometric Sequence

$$ a_1 = a_1, a_{n} = r · a_{n-1} $$

Example 1: Write an Explicit Rule for Geometric Sequence

Write the rule for the nth term for \(4, -2, 1, -\frac{1}{2}, \frac{1}{4}, …\).

Solution

The first term is 4, so \(a_1 = 4\). Divide terms to find the common ratio, r.

$$ -2 ÷ 4 = \color{red}{-\frac{1}{2}}; 1 ÷ -2 = \color{red}{-\frac{1}{2}}; -\frac{1}{2} ÷ 1 = \color{red}{-\frac{1}{2}}; \frac{1}{4} ÷ -\frac{1}{2} = \color{red}{-\frac{1}{2}} $$

It appears that the common ratio is \(-\frac{1}{2}\), so \(r = -\frac{1}{2}\).

$$ a_n = a_1 · r^{n - 1} $$

$$ a_n = 4 · \left(-\frac{1}{2}\right)^{n - 1} $$

Try It 1

Write the rule for the nth term for 3, 6, 12, 24, 48, ….

Answer

\(a_n = 3·2^{n-1}\)

Example 2: Find Explicit Geometric Rule Based on Two Terms

The 3rd term of an geometric sequence is 48, and the 6th term is –3072. Write the rule for the nth term.

Solution

This gives two points \(\color{blue}{a_3 = 48}\) and \(\color{red}{a_{6} = -3072}\). Substitute each point into the formula to obtain two equations to solve for a1 and r.

$$ a_n = a_1 · r^{n - 1} $$

$$ \begin{align} \color{blue}{48} &= \color{blue}{a_1 · r^{3 - 1}} \\ \color{red}{-3072} &= \color{red}{a_1 · r^{6 - 1}} \end{align} $$

$$ \begin{align} \color{blue}{48} &= \color{blue}{a_1 · r^{2}} \\ \color{red}{-3072} &= \color{red}{a_1 · r^{5}} \end{align} $$

Because this is a nonlinear system of equations, it must be solved using substitution. Take the first equation and solve it for a1.

$$ 48 = a_1 · r^2 $$

$$ \frac{48}{r^2} = a_1 $$

Substitute this into the second equation.

$$ -3072 = a_1 · r^5 $$

$$ -3072 = \color{blue}{\frac{48}{r^2}} · r^5 $$

Simplify the powers of r. It is dividing, so subtract the exponents.

$$ -3072 = 48 · r^3 $$

$$ -64 = r^3 $$

$$ r = -4 $$

Now that r is known, substitute it back into the first equation.

$$ \frac{48}{r^2} = a_1 $$

$$ \frac{48}{\color{blue}{(-4)}^2} = a_1 $$

$$ 3 = a_1 $$

Now write the rule for the nth term.

$$ a_n = a_1 · r^{n - 1} $$

$$ a_n = 3 · \left(-4\right)^{n - 1} $$

Try It 2

The 2nd term of an geometric sequence is \(\frac{5}{2}\), and the 7th term is \(\frac{5}{64}\). Write the rule for the nth term.

Answer

\(a_n = 5\left(\frac{1}{2}\right)^{n-1}\)

Example 3: Write the Recursive Rule for an Geometric Sequence

Write the recursive rule for 130, 26, \(\frac{26}{5}\), \(\frac{26}{25}\), ….

Solution

The common ratio appears to be \(\frac{1}{5}\), and the first term is 130. \(r = \frac{1}{5}, a_1 = 130\)

$$ a_1 = a_1, a_{n} = r · a_{n-1} $$

$$ a_1 = 130, a_{n} = \frac{1}{5} · a_{n-1} $$

Try It 3

Write the recursive rule for 5, 15, 45, 135, ….

Answer

\(a_1 = 5, a_{n} = 3 · a_{n-1}\)

Geometric Series

Deriving the formula for the sum, S, of a geometric series involves some simple factoring.

$$ S = a_1 + a_1 · r + a_1 · r^2 + a_1 · r^3 + a_1 · r^4 + \cdots + a_1 · r^{n-1} $$

Multiply both sides by r.

$$ Sr = a_1 · r + a_1 · r^2 + a_1 · r^3 + a_1 · r^4 + a_1 · r^5 + \cdots + a_1 · r^{n-1} + a_1 · r^n $$

Comparing these two first equations, the right side of this equation is \(S + a_1 · r^n - a_1\).

$$ Sr = S + a_1 · r^n - a_1 $$

$$ Sr - S = a_1 · r^n - a_1 $$

Multiply both sides by –1 and rearrange.

$$ S - Sr = a_1 - a_1 · r^n $$

Factor.

$$ S(1 - r) = a_1(1 - r^n) $$

Solve for the sum S.

$$ S_n = a_1 \frac{1 - r^n}{1 - r} $$

Because \(r^∞ ≈ 0\) when \(|r| < 1\), if an infinite sum the formula becomes

$$ S_∞ = \frac{a_1}{1-r} $$

Sum of a Geometric Series

$$ S_n = a_1 \frac{1 - r^n}{1 - r} $$

Sum of an Infinite Geometric Series

$$ S_∞ = \frac{a_1}{1-r} $$

where \(|r| < 1\)

Example 4: Find the Sum of a Geometric Series

Evaluate \(\displaystyle \sum_{n=1}^{5} 3^{n-1}\).

Solution

This is an geometric series because it is exponential in the form \(a_1 · r^{n-1}\). Comparing that to the question, \(a_1 = 1\) and \(r = 3\).

$$ S_n = a_1 \frac{1 - r^n}{1 - r} $$

$$ S_5 = 1 \frac{1 - 3^5}{1 - 3} $$

$$ S_5 = 121 $$

Try It 4

Evaluate \(\displaystyle \sum_{n=1}^{10} 2\left(\frac{1}{3}\right)^{n-1}\).

Answer

\(\frac{59048}{19683}\)

Example 5: Find the Sum of an Infinite Geometric Series

Find the sum of 8 + 0.8 + 0.08 + 0.008 + ….

Solution

This is an infinite geometric series with a1 = 8 and \(r = \frac{1}{10}\).

$$ S_∞ = \frac{a_1}{1 - r} $$

$$ S_∞ = \frac{8}{1 - \frac{1}{10}} $$

$$ S_∞ = \frac{8}{\frac{9}{10}} $$

$$ S_∞ = \frac{80}{9} $$

Try It 5

Find the sum of 12 + 0.12 + 0.0012 + 0.000012 + ….

Answer

\(\frac{400}{33}\)

Example 6: Find the Sum of an Geometric Series

Evaluate \(\displaystyle \sum_{n=0}^{∞} 3\left(\frac{1}{4}\right)^n\)

Solution

There are two issues with this problem. The formula is valid for a lower limit of 1 and an exponent of n − 1. Start by fixing the first problem. Evaluate the 0 and then the rest of the sum will then start at 1.

$$ \sum_{n=0}^{∞} 3\left(\frac{1}{4}\right)^n $$

$$ \color{blue}{3\left(\frac{1}{4}\right)^0} + \sum_{n=1}^{∞} 3\left(\frac{1}{4}\right)^n $$

$$ 3 + \sum_{n=1}^{∞} 3\left(\frac{1}{4}\right)^n $$

Fix the second issue, which was making the exponent be n − 1. Exponents are subtracted when dividing with the same base, so divide the expression by the base. But, in order to not change the value of the expression, multiply by the base.

$$ 3 + \sum_{n=1}^{∞} \color{red}{\left(\frac{1}{4}\right)}3\frac{\left(\frac{1}{4}\right)^n}{\color{purple}{\left(\frac{1}{4}\right)^1}} $$

$$ 3 + \sum_{n=1}^{∞} \color{red}{\left(\frac{3}{4}\right)}\left(\frac{1}{4}\right)^{\color{purple}{n-1}} $$

Now the sum is in the correct format with \(a_1 = \frac{3}{4}\) and \(r = \frac{1}{4}\).

$$ S_∞ = \frac{a_1}{1 - r} $$

$$ 3 + \frac{\frac{3}{4}}{1 - \frac{1}{4}} $$

$$ 3 + 1 = 4 $$

Note: This is the same answer you would have gotten if you used \(a_1 = 3\) and \(r = \frac{1}{4}\) into the infinite sum formula. However, that is mathematically inappropriate because the problem does not fit the requirements of the formula. It is simply a coincidence.

Try It 6

Evaluate \(\displaystyle \sum_{n=0}^{∞} -4\left(\frac{1}{3}\right)^{n}\)

Answer

–6

Lesson Summary

Explicit Rule for Geometric Sequences

$$ a_n = a_1 · r^{n - 1} $$

where a1 is the first term and r is the common ratio

Recursive Rule for Geometric Sequence

$$ a_1 = a_1, a_{n} = r · a_{n-1} $$


Sum of a Geometric Series

$$ S_n = a_1 \frac{1 - r^n}{1 - r} $$

Sum of an Infinite Geometric Series

$$ S_∞ = \frac{a_1}{1-r} $$

where \(|r| < 1\)

Helpful videos about this lesson.

Practice Exercises

  1. Explain how to find the common ratio.

    Is the sequence geometric? If so, what is the common ratio?

  2. \(3, -\frac{6}{5}, \frac{12}{25}, -\frac{24}{125}, …\)
  3. \(2, 12, 72, 432, …\)
  4. Write the rule for the sequence. (a) explicit (b) recursive

  5. \(\frac{3}{2}, 1, \frac{2}{3}, \frac{4}{9}, …\)
  6. –7, 14, –28, 56, …
  7. 8, 20, 50, 125, …
  8. Write the rule for the nth term given the following two terms.

  9. \(a_3 = \frac{4}{3}, a_8 = \frac{128}{729}\)
  10. \(a_4 = -50, a_{11} = -3906250\)
  11. Evaluate the sum.

  12. 5 + 10 + 20 + 40 + 80 + … + 640
  13. \(\frac{5}{3} + \frac{5}{2} + \frac{15}{4} + \frac{45}{8} + … + \frac{405}{32}\)
  14. \(\displaystyle \sum_{i=1}^{7} \frac{1}{2}(3)^{i-1}\)
  15. \(\displaystyle \sum_{k=1}^{5} 2\left(\frac{5}{2}\right)^{k-1}\)
  16. \(\displaystyle \sum_{n=1}^{10} 10\left(\frac{1}{5}\right)^{n-1}\)
  17. \(\displaystyle \sum_{n=1}^{∞} 6\left(\frac{1}{4}\right)^{n-1}\)
  18. \(\displaystyle \sum_{n=0}^{∞} -4\left(\frac{2}{3}\right)^{n}\)
  19. Problem Solving

  20. A Christian wants to spread her love for Jesus to others. She tells three people about Jesus who then they each tell three other people each. If each person tells three people about Jesus who then tell three more people. How many total people have been told about Jesus after the 10th set of people have been told about Him?
  21. Mixed Review

  22. (10-03) Is the sequence given by the rule \(a_n = 3n + 5\) geometric, arithmetic, or neither?
  23. (10-03) Evaluate \(\displaystyle \sum_{n=1}^{15} 2n - 4\).
  24. (10-02) Write the series in sigma notation: 2, 6, 18, 54, …, 486.
  25. (10-01) Write the first five terms of the sequence \(a_n = 3n + 1\).

Answers

  1. Take a term and divide it by the previous term.
  2. Yes; \(-\frac{2}{5}\)
  3. Yes; 6
  4. \(a_n = \frac{3}{2}\left(\frac{2}{3}\right)^{n-1}; a_1 = \frac{3}{2}, a_{n} = \frac{2}{3} a_{n-1}\)
  5. \(a_n = -7\left(-2\right)^{n-1}; a_1 = -7, a_{n} = -2 a_{n-1}\)
  6. \(a_n = 8\left(\frac{5}{2}\right)^{n-1}; a_1 = 8, a_{n} = \frac{5}{2} a_{n-1}\)
  7. \(a_n = 3\left(\frac{2}{3}\right)^{n-1}\)
  8. \(a_n = -\frac{2}{5}\left(5\right)^{n-1}\)
  9. 1275
  10. about 34.635
  11. \(\frac{1093}{2}\)
  12. \(\frac{1031}{8}\)
  13. about 12.500
  14. 8
  15. –12
  16. 88572
  17. Arithmetic (because it is linear)
  18. 180
  19. \(\displaystyle \sum_{n=1}^{6} 2(3)^{n-1}\)
  20. 4, 7, 10, 13, 16