Precalculus by Richard Wright

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

Summary: In this section, you will:

- Write the explicit rule for an geometric sequence.
- Write the recursive rule for an geometric sequence.
- Evaluate the sum for an geometric series.

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

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 10^{th} 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.

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^{1}, 1·3^{2}, 1·3^{3}, 1·3^{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 *n*^{th} term.

*a*_{n} = *a*_{1} · *r*^{n−1}

*a*_{n} = *a*_{1} · *r*^{n−1}

where *a*_{1} is the first term and *r* is the common ratio

*a*_{1} = *a*_{1}, *a*_{n} = *r* · *a*_{n−1}

Write the rule for the *n*^{th} 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}

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

Write the rule for the *n*^{th} term for 3, 6, 12, 24, 48, ….

**Answer**

*a*_{n} = 3 · 2^{n−1}

The 3^{rd} term of an geometric sequence is 48, and the 6th term is –3072. Write the rule for the *n*^{th} term.

**Solution**

This gives two points *a*_{3} = 48 and *a*_{6} = −3072. Substitute each point into the formula to obtain two equations to solve for *a*_{1} 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 *a*_{1}.

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 *n*^{th} term.

*a*_{n} = *a*_{1} · *r*^{n−1}

*a*_{n} = 3 · (−4)^{n−1}

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

**Answer**

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

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}

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

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

**Answer**

*a*_{1} = 5, *a*_{n} = 3 · *a*_{n−1}

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} + ⋯ + *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} + ⋯ + *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

$$ S_∞ = \frac{a_1}{1-r} $$

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

$$ S_∞ = \frac{a_1}{1-r} $$

where |*r*| < 1

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

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

**Answer**

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

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

**Solution**

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

$$ S_∞ = \frac{a_1}{1 - r} $$

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

$$ S_∞ = \frac{8}{\frac{9}{10}} $$

$$ S_∞ = \frac{80}{9} $$

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

**Answer**

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

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.

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

**Answer**

–6

*a*_{n} = *a*_{1} · *r*^{n−1}

where *a*_{1} is the first term and *r* is the common ratio

*a*_{1} = *a*_{1}, *a*_{n} = *r* · *a*_{n−1}

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

$$ S_∞ = \frac{a_1}{1-r} $$

where |*r*| < 1

Helpful videos about this lesson.

- Mr. Wright Teaches the Lesson (https://youtu.be/wkchAUWMdiY)

- Geometric Sequences (http://openstaxcollege.org/l/geometricseq)
- Determine the Type of Sequence (http://openstaxcollege.org/l/sequencetype)
- Geometric Series (http://openstaxcollege.org/l/geometricser)

- Explain how to find the common ratio.
- Is the sequence geometric? If so, what is the common ratio?
- \(3, -\frac{6}{5}, \frac{12}{25}, -\frac{24}{125}, …\)
- 2, 12, 72, 432, …
- Write the rule for the sequence. (a) explicit (b) recursive
- \(\frac{3}{2}, 1, \frac{2}{3}, \frac{4}{9}, …\)
- –7, 14, –28, 56, …
- 8, 20, 50, 125, …
- Write the rule for the
*n*^{th}term given the following two terms. - \(a_3 = \frac{4}{3}, a_8 = \frac{128}{729}\)
*a*_{4}= −50,*a*_{11}= −3906250- Evaluate the sum.
- 5 + 10 + 20 + 40 + 80 + … + 640
- \(\frac{5}{3} + \frac{5}{2} + \frac{15}{4} + \frac{45}{8} + … + \frac{405}{32}\)
- \(\displaystyle \sum_{i=1}^{7} \frac{1}{2}(3)^{i-1}\)
- \(\displaystyle \sum_{k=1}^{5} 2\left(\frac{5}{2}\right)^{k-1}\)
- \(\displaystyle \sum_{n=1}^{10} 10\left(\frac{1}{5}\right)^{n-1}\)
- \(\displaystyle \sum_{n=1}^{∞} 6\left(\frac{1}{4}\right)^{n-1}\)
- \(\displaystyle \sum_{n=0}^{∞} -4\left(\frac{2}{3}\right)^{n}\)
- Problem Solving
- 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 10
^{th}set of people have been told about Him? - Mixed Review
- (10-03) Is the sequence given by the rule
*a*_{n}= 3*n*+ 5 geometric, arithmetic, or neither? - (10-03) Evaluate \(\displaystyle \sum_{n=1}^{15} 2n - 4\).
- (10-02) Write the series in sigma notation: 2, 6, 18, 54, …, 486.
- (10-01) Write the first five terms of the sequence
*a*_{n}= 3*n*+ 1.

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