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7. Sequences
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Integrated Math 1

7.03 Exponential functions as geometric sequences

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Concept summary

In addition to the arithmetic sequence, there is another special type of sequence called a geometric sequence. These sequences have a constant multiplicative pattern.

Geometric sequence

A sequence of numbers in which each consecutive pair of numbers has a common ratio.

Example:

1, 3, 9, 27, 81,\ldots

Common ratio

The ratio of consecutive terms in a geometric sequence: r=\dfrac{a_n}{a_{n-1}}

Example:

1, 3, 9, 27, 81,\ldots has r=3

The nth term, a_n, of a geometric sequence is given by the explicit formula or general rule:

\displaystyle a_n=a_1 r^{n-1}
\bm{a_1}
The first term
\bm{r}
The common ratio

For a geometric sequence, we can also use a recursive formula and the first term to describe the sequence:

\displaystyle a_n=ra_{n-1} \text{ and given } a_1
\bm{a_n}
The current term
\bm{a_{n-1}}
The previous term
\bm{r}
The common ratio
\bm{a}
The value of the first term

Notations other than a_n such as t_n, T_n, b_n, u_n, ... can be used for different contexts.

When we represent an geometric sequence as a exponential function whose domain is a subset of the integers, we generally use function notation and simplify:

\displaystyle a(n)=a(1)\cdot r^{n-1}
\bm{a(n)}
The value of the nth term
\bm{a(1)}
The first term
\bm{r}
The common ratio

Worked examples

Example 1

Identify whether each sequence is geometric or not. If it is a geometric sequence, write the explicit rule.

a

3.1,\,5.6,\,8.1,\,10.6,\, \ldots

Approach

The ratio of consecutive terms in a geometric sequence is always the same. That is, if the sequence a_1,\,a_2,\,a_3,\,a_4,\,\ldots is geometric, then \dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=\ldots

Solution

From the given sequence: a_1=3.1,\,a_2=5.6,\,a_3=8.1,\,a_4=10.6,\, \ldots

Observe that \dfrac{a_2}{a_1}=\dfrac{5.6}{3.1}=1.81 and \dfrac{a_3}{a_2}=\dfrac{8.1}{5.6}=1.45. So we have \dfrac{a_2}{a_1} \neq\dfrac{a_3}{a_2}. Because the ratio of consecutive terms for this sequence is not always the same this sequence is not geometric.

Reflection

Notice that a_2-a_1=5.6-3.1=2.5,\,a_3-a_2=8.1-5.6=2.5,\,a_4-a_3=10.6-8.1=2.5. The sequence has a common difference of 2.5. This tells us that the sequence is actually an arithmetic sequence.

b

6,\,18,\, 54,\,162,\, \ldots

Approach

If the sequence a_1,\,a_2,\,a_3,\,a_4,\,\ldots is geometric, then \dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=\ldots

The explicit rule for the geometric sequence is given by a_n = a_1 r^{n - 1} where a_1 is the first term in the sequence and r is the common ratio.

Solution

From the given sequence: a_1=6,\,a_2=18,\,a_3=54,\,a_4=162,\, \ldots

Check each ratio: \dfrac{a_2}{a_1}=\dfrac{18}{6}=3,\quad\dfrac{a_3}{a_2}=\dfrac{54}{18}=3,\quad\dfrac{a_4}{a_3}=\dfrac{162}{54}=3

This sequence is geometric since each ratio is equal.

To write the explicit rule we can substitute a_1=6 and r=3 into a_n = a_1 r^{n - 1} which gives us a_n = 6(3)^{n - 1}

Reflection

Use parentheses around the common ratio to separate it from the first term: 6(3)^{n-1} vs. 63^{n-1}

c

-4,\,20,\, -100,\,500,\, \ldots

Solution

From the given sequence: a_1=-4,\,a_2=20,\,a_3=-100,\,a_4=500,\, \ldots

Check each ratio: \dfrac{a_2}{a_1}=\dfrac{20}{-4}=-5,\quad\dfrac{a_3}{a_2}=\dfrac{-100}{20}=-5,\quad\dfrac{a_4}{a_3}=\dfrac{500}{-100}=-5

This sequence is geometric since each ratio is equal.

To write the explicit rule we can substitute a_1=-4 and r=-5 into a_n = a_1 r^{n - 1} which gives us a_n = -4(-5)^{n - 1}

Reflection

When the common ratio is negative, there is not a single related exponential function with a restricted domain as the graph alternates back and forth above and below the x-axis which exponential models do not do.

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Example 2

Consider the geometric sequence defined by: a_n=64 \left(\frac{1}{2} \right)^{n-1}

a

Write the recursive formula for this geometric sequence.

Approach

We can use the explicit rule for a geometric sequence a_n=a_1r^{n-1} to identify the first term and common ratio and then substitute those into the recursive formula.

Solution

From the equation we see that: a_1=64 and r=\dfrac{1}{2}

Substituting into the recursive formula we get: a_n=\dfrac{1}{2}a_{n-1} and a_1=64

Reflection

Without stating the first term, we cannot uniquely identify a sequence, if we just say the recursive formula is a_n=\dfrac{1}{2}a_{n-1}, this describes any sequence with common ratio of \dfrac{1}{2}, not necessarily the one described by the explicit formula.

b

Find a_{10}.

Approach

We can use our explicit rule to find the tenth term, a_{10}.

Solution

\displaystyle a_n\displaystyle =\displaystyle 64\left(\frac{1}{2}\right)^{n-1}Explicit rule
\displaystyle a_{10}\displaystyle =\displaystyle 64\left(\frac{1}{2}\right)^{(10-1)}Substitute 10 for n
\displaystyle a_{10}\displaystyle =\displaystyle 64\left(\frac{1}{2}\right)^{9}Evaluate the difference
\displaystyle a_{10}\displaystyle =\displaystyle \dfrac{1}{8}Evaluate the expression

Reflection

If we wanted, we could also continue the pattern using the recursive rule in a table to find the tenth term:

n12345678910
a_n6432168421\dfrac{1}{2}\dfrac{1}{4}\dfrac{1}{8}

Example 3

Consider the first four terms of a geometric sequence which have been plotted on the coordinate plane:

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a_n
a

Identify the first term and common ratio.

Approach

The first term is a_1, so we can go to n=1 on the horizontal axis to find the value from the vertical axis.

To find the common ratio, we can find \dfrac{a_2}{a_1} or any other consecutive pair of values.

Solution

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From the graph we can find that a_1=3.

We can also see that a_2=6,a_3=12, and {a_4=24} so we can calculate the common ratio by finding the ratio of \dfrac{a_2}{a_1}.

\dfrac{6}{3}=2 so have r=2.

The first term is a_1=3 and the common ratio is r=2.

Reflection

We can use any pair of consecutive points to find the common ratio:

\frac{a_3}{a_2}=\frac{12}{6}=2

\frac{a_4}{a_3}=\frac{24}{12}=2

If it is hard to read some points due to scale, we could work backwards to identify the first term, by dividing larger terms by the common ratio repeatedly.

In this case, starting with a_4=24, we would need to divide three times: \dfrac{24}{2}=12, then \dfrac{12}{2}=6, then \dfrac{6}{2}=3.

b

Write an explicit rule using function notation to represent the geometric sequence as an exponential function, a(n).

Approach

The general rule will start in the form a(n)=a(1)\cdot r^{n-1}, where a(1) represents the first term, so we need to find a(1) and r and then substitute and simplify.

Solution

We already know both of these values from part (a), so we just need to substitute them in and simplify.

Equation in function notation: a(n)=3\cdot 2^{n-1} or fully simplified a(n)=\dfrac{3}{2}\cdot 2^n.

Reflection

The simplified version a(n)=\dfrac{3}{2} \cdot 2^n we can see the growth factor for the related exponential function is 2 and the value of the y-intercept is y=\dfrac{3}{2}. This form can be particularly helpful for connecting the geometric sequence to an exponential model.

c

Describe the domain of the exponential function that is related to the sequence.

Approach

We will assume that the first term given has n=1 and can see which values of n are graphed.

Solution

The domain is all positive integers, which can be written as n \in \{1, 2, 3, 4, \ldots \} or more precisely as n \in \Z^+.

Reflection

Sometimes we will be given a sequence with a 0th term, in which case we would need to include 0 in the domain.

This could be represented by giving a_0 in the recursive formula or having a point at n=0 on the graph.

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Outcomes

M1.F.BF.A.2

Define sequences as functions, including recursive definitions, whose domain is a subset of the integers. Write explicit and recursive formulas for arithmetic and geometric sequences in context and connect them to linear and exponential functions.*

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

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