# 8.3: Geometric Sequences

- Page ID
- 83160

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Geometric sequences have a common ratio. Each term after the first term is obtained by multiplying the previous term by \(r\), the common ratio. As an example, the following sequence does not have a common difference, so it is not an arithmetic sequence. Instead, this sequence has a **common ratio**, \(r\):

\[\dfrac{a_n}{a_{n−1}} = r = 2 \nonumber\]

and it is a geometric sequence:

\(2,4,8,16,32,64,128\)

Notice that each term is double the previous term. By multiplying any term by \(2\), we obtain the subsequent term. A common ratio is the hallmark signature of a geometric sequence.

If the sequence: \(a_1 , a_2, a_3, a_4 , … , a_{n−1}, a_n, …\) exhibits a pattern (\(a ≠ 0\) and \(r ≠ 0\)) such that

\[a_1, a_1r, a_1r^2 , a_1r^3 , … , a_1r^{n−1} , a_1r^n , …\]

Then the sequence is geometric and \(r\) is called the common ratio, where \(\dfrac{a_n}{a_{n-1}} = r\)

A geometric sequence is analogous to an exponential function, \(f(x) = ab^x\), where \(a\) and \(b\) are constants, \(a=\) any real number and \(b > 0\). The general term \(a_n\) for a geometric sequence will mimic the exponential function formula, but modified in the following way:

- Instead of \(x =\) any real number, the domain of the geometric sequence function is the set of natural numbers \(n\).
- The constant \(a\) will become the first term, or \(a_1\), of the geometric sequence.
- The constant \(b\) is replaced by the common ratio \(r\), but \(r\) can be positive or negative.

The geometric sequence with common ratio \(r\):

\[a_1, a_1r, a_1r^2 , a_1r^3 , … , a_1r^{n−1} , …\]

has general term

\[f(n) = a_n = a_1r^{n−1}\]

The first term of the geometric sequence is \(a_1\), or \(a_1r^0\). Recall that \(r^0 = 1\).It’s worth mentioning that, in some cases, the first term is better notated as \(a_0\) rather than \(a_1\). If we use \(a_0 =\) first term (starting the sequence at \(n = 0\)), then the geometric sequence would be notated: \(a_0, a_0r, a_0r^2 , a_0r^3\), … and the general term is \(a_n = a_0r^n\). Although the term-number no longer matches the subscript (i.e. \(a_1=\) second term, \(a_2 =\) third term, etc.), the exponent on \(r\) tells us how many times \(r\) was applied. In real-life problems that have an initial value to which \(r\) is repetitively multiplied, allow yourself the flexibility to call the initial amount \(a_0\).

Determine the common ratio of the geometric sequence: \(15, 45, 135, 405, …\) and give the general term, \(a_n\). Then find the \(10^{\text{th}}\) term of the sequence, or \(a_{10}\).

**Solution**

Finding the common ratio is a matter of dividing any term by its previous term:

\(\dfrac{45}{15} = 3 = r\).

Therefore, the general term of the sequence is:

\(a_n = 15 \cdot 3^{n-1}\)

The general term gives us a formula to find \(a_{10}\). Plug \(n = 10\) into the general term \(a_n\).

\(a_{10} = 15 \cdot 3^{10−1} = 15 \cdot 3^9 = 295245\)

Determine the common ratio of the geometric sequence: \(8, −12, 18, −27, …\) and give the general term \(a_n\). Then find the \(7^{\text{th}}\) term of the sequence.

**Solution**

Notice the sequence alternates in sign value: positive, negative, positive, negative, … An alternating sequence occurs when \(r < 0\). We expect the \(r\)-value to be negative.

The common ratio is found by dividing two consecutive terms. Let’s divide \(\dfrac{a_2}{a_1}\).

\(\dfrac{−12}{8} = -\dfrac{3}{2} = -1.5 = r\)

Therefore, the general term of the sequence is:

\(a_n = -1.5 \cdot 8^{n-1}\)

The general term gives us a formula to find \(a_7\). Plug into \(n = 7\) in \(a_n\) to find the \(7^{\text{th}}\) term:

\(a_7 = 8(−1.5)^{7−1} = 8(−1.5)^6 = 91.125\)

A filtering process can reduce Chemical B by \(10\%\). The process can be repeated and have the same reduction rate of \(10\%\) each time. Initially, there is \(7\) mg of Chemical B before filtering. How much of Chemical B remains after \(4\) filtering processes? Round the answer to \(2\) decimal places.

**Solution**

If chemical B is reduced by \(10\%\), then \(90\%\) remains after filtering. The sequence would end after \(4\) filters. The common ratio \(r = 0.9\) and we apply this common ratio \(4\) times to the initial value, \(7\) mg:

Rather than performing each multiplication separately, it’s easier to compute the remaining quantity of Chemical B using the formula for the general term:

\(7 \cdot (0.9)^4 = 4.5927\)

**Answer** After \(4\) filtering processes, \(4.59\) mg of Chemical B remains.

The average annual inflation was \(1.75\%\) between 2011 and 2020. The cost of a Big Mac rose with inflation. If inflation was the only factor for an increase in the price of a Big Mac, how much would a Big Mac cost in 2020 if nine years earlier, in 2011, it cost \($4.20\)?

**Solution**

The common ratio is a value greater than one. Each year, inflation drove up the price of a Big Mac so that consumers were paying \(100 \% + 1.75\% = 101.75\%\) for the burger over the previous year’s cost. The common ratio is the percentage as a decimal. \(r = 1.0175\).

\(\begin{array} &&4.20(1.0175)^9 ≈ 4.91 & a_0 = 4.20. \text{Apply the common ratio to each year 2011-2020.} \end{array}\)

**Answer** The cost of a Big Mac would have cost \($4.91\) in 2020.

## Percentage Increase or Decrease

As demonstrated in Example \(8.3.3\) (percentage decrease) and Example \(8.3.4\) (percentage increase), if the change to the initial quantity is given as a constant percentage (the percentage itself doesn’t change), the sequence will be geometric. The sequence is also sometimes called a geometric progression.

The geometric progression in Example \(8.3.3\) is a decreasing sequence. The \(r\)−value is calculated by considering the offset from \(100\%\). If the percentage, \(p\%\), is a decrease, subtract the percentage from \(100\%\): \((100\% − p\%)\) then change the value to a decimal by moving the decimal two places left (or divide by \(100\)). If the sequence decreases, \(0 < r < 1\).The geometric progression in Example \(8.3.4\) is an increasing sequence. The \(r\)−value is calculated by considering the offset from \(100\%\). If the percentage, \(p\%\), is an increase, add the percentage to \(100\%\): \((100\% + p\%)\) then change the value to a decimal by moving the decimal two places left (or divide by \(100\)). If the sequence increases, \(r > 1\).

Alice puts \(1\) grain of rice on the first square of an \(8 \times 8\) chess board. The White Rabbit tells her to put double the amount of rice in each successive square thereafter. The Cheshire Cat chuckles and tells Alice, “you should join the Mad Hatters if you think you can accomplish this task.” How many grains of rice would be on the \(64^{\text{th}}\) square?

**Solution**

If one grain of rice occupies the first square, \(2\) the second square, \(4\) the third square, then the common ratio is \(r = 2\), doubling the grains of rice \(63\) times. Remember, we don’t start doubling until the \(2^{\text{nd}}\) square, so although there are \(64\) squares, we double the quantity of rice \(63\) times.

\(1 \cdot 2^{63} ≈ 9 x 10^{18}\) grains of rice on the \(64^{\text{th}}\) square!

It’s hard to fathom such a large quantity!

## Try It! (Exercises)

For #1-6, the sequence is either arithmetic, geometric, or neither. If arithmetic, give the common difference. If geometric, give the common ratio. If neither, show how it fails to have a common difference or a common ratio.

- \(100, 200, 300, 400, …\)
- \(10, 100, 103 , 104 , …\)
- \(1 2 , 1 4 , 1 8 , 1 16 , …\)
- \(1 2 , 2 3 , 3 4 , 4 5 , . ..\)
- \(8, −6, 4.5, −3.375, …\)
- \(1, −3, −7, −11, …\)

For #7-12, the general term of the sequence is given.

- Is the sequence geometric or arithmetic?
- State the first five terms, starting with \(n = 1\).
- Find the value \(a_9\).

- \(a_n = 6 \cdot 2^{n−1}\)
- \(a_n = 52 − 13n\)
- \(a_n = 5n\)
- \(a_n = 5 \left( \dfrac{2}{3} \right)^n\)
- \(a_n = 2 + \dfrac{n}{4}\)
- \(a_n = 3(−1)^n\)

For #13-15, a term of a geometric sequence is given along with the common ratio \(r\). Find the first \(5\) terms of the geometric sequence, starting with \(n = 1\), and state the general term, \(a_n\).

- \(a_1 = 2.5, r = 4\)
- \(a_3 = 162, r = 3\)
- \(a_5 = \dfrac{1}{8} , r = \dfrac{1}{2}\)

For #16-18, two terms of a geometric sequence are given. Find the first \(5\) terms of the geometric sequence, starting with \(n = 1\), and state the general term, \(a_n\).

- \(a_2 = 3, a_3 = \dfrac{9}{10}\)
- \(a_1 = \dfrac{2}{5} , a_4 = −\dfrac{16}{5}\)
- \(a_2 = 40, a_4 = \dfrac{32}{5}\)
- A solar company guarantees that the solar panels installed on a home can produce \(3500\) kWh for the first year. Every year thereafter, the guaranteed annual production (in kWh) declines by \(2\%\). How many annual kWh of energy would be guaranteed \(8\) years after installation? Round answer to the nearest whole kWh.
- Global carbon emissions have risen by approximately \(2.6\%\) annually from 1960 to 2010. If \(2500\) million metric tons of carbon were emitted in 1960, how many metric tons of carbon were emitted in 2010? Round the answer to the nearest whole million metric ton.
- When the IRS makes a mistake in your favor, the IRS must pay you back, plus interest, on the overpayment. How much, in total, would the IRS owe you if you overpaid \($1,500\) to the IRS and that amount earned interest over \(3\) years at a constant rate of \(3\%\) annual interest? Round your answer to the nearest cent.
- If there are \(30\) years in a generation, how many direct ancestors did each of us have \(150\) years ago? By direct ancestor we mean birth parents, grandparents, great-grandparents, and so on.
- Organisms in nature draw Carbon-14 from the environment, but once the organism dies, it starts to lose carbon-14 at an exponential rate. Every 5730 years, the amount Carbon-14 is cut in half. Carbon dating can be used on samples of bone, cloth, wood, and plant fibers.
- How many years of decay before \(\dfrac{1}{4}\) remains?
- How many years of decay before \(\dfrac{1}{8}\) remains?
- How many years of decay before \(\dfrac{1}{2^6}\) remains?

- If the midpoints of the sides of an equilateral triangle are joined by straight lines, the new figure will be an equilateral triangle with a perimeter equal to half the original. If we start with an equilateral triangle with perimeter \(3\) cm, what is the perimeter of the \(5^{\text{th}}\) “nested” triangle, as described?
- Reservoirs can be the source of water supply for millions of people. Changes can occur to any water supply due to inflow and outflow, but evaporation is one of the factors of water depletion. Suppose a reservoir contains an average of \(1.4\) billion gallons of water and loses water due to evaporation at a rate of \(2\%\) per month. Without considering any other changes to the reservoir’s volume, how much water will have evaporated over a one-year period?
- Zeno’s Paradox is an observation which seems absurd, yet it starts sounding logically acceptable in relation to geometric sequences! Zeno’s Paradox reads:

Suppose Atalanta wishes to walk to the end of a path. Before she can get there, she must get halfway there. Before she can get halfway there, she must get a quarter of the way there. Before traveling a quarter, she must travel one-eighth; before an eighth, one-sixteenth; and so on.

Zeno’s paradox questions the conclusion of a geometric sequence, which paradoxically questions Atalanta’s ability to complete her walk to the end of the path! Our brain battles the fact that the sequence is infinite against our observable experience – of course Atalanta can walk to the end of the path! A related paradox to ponder: when would you say that the perimeter of a nested triangle in Problem #24 is equal to zero? This question might seem absurd, just like Zeno’s Paradox! Use your own thoughts to contemplate the question and debate your conclusion with a logical argument.