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Geometric Sequences

June 23, 2024 | by Bloom Code Studio

Learning Objectives

In this section, you will:

  • Find the common ratio for a geometric sequence.
  • List the terms of a geometric sequence.
  • Use a recursive formula for a geometric sequence.
  • Use an explicit formula for a geometric sequence.

Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.

Finding Common Ratios

The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.A sequence , {1, 6, 36, 216, 1296, ...} that shows all the numbers have a common ratio of 6.

DEFINITION OF A GEOMETRIC SEQUENCE

geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a1π‘Ž1 is the initial term of a geometric sequence and rπ‘Ÿ is the common ratio, the sequence will be

{a1, a1r,a1r2,a1r3,…}.{π‘Ž1, π‘Ž1π‘Ÿ,π‘Ž1π‘Ÿ2,π‘Ž1π‘Ÿ3,…}.

HOW TO

Given a set of numbers, determine if they represent a geometric sequence.

  1. Divide each term by the previous term.
  2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.

EXAMPLE 1

Finding Common Ratios

Is the sequence geometric? If so, find the common ratio.

  1. ⓐ 1,2,4,8,16,…1,2,4,8,16,…
  2. β“‘Β 48,12,4,Β 2,…48,12,4,Β 2,…

Solution

Divide each term by the previous term to determine whether a common ratio exists.

  1. ⓐ 21=242=284=2168=221=242=284=2168=2The sequence is geometric because there is a common ratio. The common ratio is 2.
  2. β“‘Β 1248=14412=1324=121248=14412=1324=12The sequence is not geometric because there is not a common ratio.

Analysis

The graph of each sequence is shown inΒ Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.

Graph of two sequences where graph (a) is geometric and graph (b) is exponential.

Figure 1

Q&A

If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?

No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.

TRY IT #1

Is the sequence geometric? If so, find the common ratio.

5,10,15,20,…5,10,15,20,…

TRY IT #2

Is the sequence geometric? If so, find the common ratio.

100,20,4,45,…100,20,4,45,…

Writing Terms of Geometric Sequences

Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a1=βˆ’2π‘Ž1=βˆ’2 and the common ratio is r=4,π‘Ÿ=4, we can find subsequent terms by multiplying βˆ’2β‹…4βˆ’2β‹…4 to get βˆ’8βˆ’8 then multiplying the result βˆ’8β‹…4βˆ’8β‹…4 to get βˆ’32βˆ’32 and so on.

a1=βˆ’2a2=(βˆ’2β‹…4)=βˆ’8a3=(βˆ’8β‹…4)=βˆ’32a4=(βˆ’32β‹…4)=βˆ’128π‘Ž1=βˆ’2π‘Ž2=(βˆ’2β‹…4)=βˆ’8π‘Ž3=(βˆ’8β‹…4)=βˆ’32π‘Ž4=(βˆ’32β‹…4)=βˆ’128

The first four terms are {–2, β€“8, β€“32, β€“128}.{–2, β€“8, β€“32, β€“128}.

HOW TO

Given the first term and the common factor, find the first four terms of a geometric sequence.

  1. Multiply the initial term,Β a1,π‘Ž1,Β by the common ratio to find the next term,Β a2.π‘Ž2.
  2. Repeat the process, usingΒ an=a2π‘Žπ‘›=π‘Ž2Β to findΒ a3π‘Ž3Β and thenΒ a3π‘Ž3Β to findΒ a4,π‘Ž4,Β until all four terms have been identified.
  3. Write the terms separated by commons within brackets.

EXAMPLE 2

Writing the Terms of a Geometric Sequence

List the first four terms of the geometric sequence with a1=5π‘Ž1=5 and r=–2.π‘Ÿ=–2.

Solution

Multiply a1π‘Ž1 by βˆ’2βˆ’2 to find a2.π‘Ž2. Repeat the process, using a2π‘Ž2 to find a3,π‘Ž3, and so on.

a1=5a2=βˆ’2a1=βˆ’10a3=βˆ’2a2=20a4=βˆ’2a3=βˆ’40π‘Ž1=5π‘Ž2=βˆ’2π‘Ž1=βˆ’10π‘Ž3=βˆ’2π‘Ž2=20π‘Ž4=βˆ’2π‘Ž3=βˆ’40

The first four terms are {5,–10,20,–40}.{5,–10,20,–40}.

TRY IT #3

List the first five terms of the geometric sequence with a1=18π‘Ž1=18 and r=13.π‘Ÿ=13.

Using Recursive Formulas for Geometric Sequences

A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.

RECURSIVE FORMULA FOR A GEOMETRIC SEQUENCE

The recursive formula for a geometric sequence with common ratio rπ‘Ÿ and first term a1π‘Ž1 is

an=ranβˆ’1,nβ‰₯2π‘Žπ‘›=π‘Ÿπ‘Žπ‘›βˆ’1,𝑛β‰₯2

HOW TO

Given the first several terms of a geometric sequence, write its recursive formula.

  1. State the initial term.
  2. Find the common ratio by dividing any term by the preceding term.
  3. Substitute the common ratio into the recursive formula for a geometric sequence.

EXAMPLE 3

Using Recursive Formulas for Geometric Sequences

Write a recursive formula for the following geometric sequence.

{6, 9, 13.5, 20.25, …}{6, 9, 13.5, 20.25, …}

Solution

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

r=96=1.5π‘Ÿ=96=1.5

Substitute the common ratio into the recursive formula for geometric sequences and define a1.π‘Ž1.

an=ranβˆ’1an=1.5anβˆ’1for nβ‰₯2a1=6π‘Žπ‘›=π‘Ÿπ‘Žπ‘›βˆ’1π‘Žπ‘›=1.5π‘Žπ‘›βˆ’1for π‘›β‰₯2π‘Ž1=6

Analysis

The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown inΒ Figure 2

Graph of the geometric sequence.

Figure 2

Q&A

Do we have to divide the second term by the first term to find the common ratio?

No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.

TRY IT #4

Write a recursive formula for the following geometric sequence.

{2, 43, 89, 1627, …}{2, 43, 89, 1627, …}

Using Explicit Formulas for Geometric Sequences

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.

an=a1rnβˆ’1π‘Žπ‘›=π‘Ž1π‘Ÿπ‘›βˆ’1

Let’s take a look at the sequence {18, 36, 72, 144, 288, …}.{18, 36, 72, 144, 288, …}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is

an=18β‹…2nβˆ’1π‘Žπ‘›=18Β·2π‘›βˆ’1

The graph of the sequence is shown inΒ Figure 3.

Graph of the geometric sequence.

Figure 3

EXPLICIT FORMULA FOR A GEOMETRIC SEQUENCE

The n𝑛 th term of a geometric sequence is given by the explicit formula:

an=a1rnβˆ’1π‘Žπ‘›=π‘Ž1π‘Ÿπ‘›βˆ’1

EXAMPLE 4

Writing Terms of Geometric Sequences Using the Explicit Formula

Given a geometric sequence with a1=3π‘Ž1=3 and a4=24,π‘Ž4=24, find a2.π‘Ž2.

Solution

The sequence can be written in terms of the initial term and the common ratio r.π‘Ÿ.

3,3r,3r2,3r3,…3,3π‘Ÿ,3π‘Ÿ2,3π‘Ÿ3,…

Find the common ratio using the given fourth term.

an=a1rnβˆ’1a4=3r324=3r38=r3r=2Write the fourth term of sequence in terms of Ξ±1and rSubstitute 24fora4DivideSolve for the common ratioπ‘Žπ‘›=π‘Ž1π‘Ÿπ‘›βˆ’1π‘Ž4=3π‘Ÿ3Write the fourth term of sequence in terms of π›Ό1and π‘Ÿ24=3π‘Ÿ3Substitute 24forπ‘Ž48=π‘Ÿ3Divideπ‘Ÿ=2Solve for the common ratio

Find the second term by multiplying the first term by the common ratio.

a2=2a1=2(3)=6π‘Ž2=2π‘Ž1=2(3)=6

Analysis

The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.

TRY IT #5

Given a geometric sequence with a2=4π‘Ž2=4 and a3=32π‘Ž3=32 , find a6.π‘Ž6.

EXAMPLE 5

Writing an Explicit Formula for the n𝑛 th Term of a Geometric Sequence

Write an explicit formula for the nth𝑛th term of the following geometric sequence.

{2, 10, 50, 250, …}{2, 10, 50, 250, …}

Solution

The first term is 2. The common ratio can be found by dividing the second term by the first term.

102=5102=5

The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.

an=a1r(nβˆ’1)an=2β‹…5nβˆ’1π‘Žπ‘›=π‘Ž1π‘Ÿ(π‘›βˆ’1)π‘Žπ‘›=2β‹…5π‘›βˆ’1

The graph of this sequence inΒ Figure 4Β shows an exponential pattern.

Graph of the geometric sequence.

Figure 4

TRY IT #6

Write an explicit formula for the following geometric sequence.

{–1, 3, β€“9, 27, …}{–1, 3, β€“9, 27, …}

Solving Application Problems with Geometric Sequences

In real-world scenarios involving geometric sequences, we may need to use an initial term of a0π‘Ž0 instead of a1.π‘Ž1. In these problems, we can alter the explicit formula slightly by using the following formula:

an=a0rnπ‘Žπ‘›=π‘Ž0π‘Ÿπ‘›

EXAMPLE 6

Solving Application Problems with Geometric Sequences

In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.

  1. ⓐWrite a formula for the student population.
  2. β“‘Estimate the student population in 2020.

Solution

  1. ⓐThe situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.LetΒ P𝑃 be the student population andΒ n𝑛 be the number of years after 2013. Using the explicit formula for a geometric sequence we getPnΒ =284β‹…1.04n𝑃𝑛 =284β‹…1.04𝑛
  2. β“‘We can find the number of years since 2013 by subtracting.2020βˆ’2013=72020βˆ’2013=7We are looking for the population after 7 years. We can substitute 7 forΒ n𝑛 to estimate the population in 2020.P7=284β‹…1.047β‰ˆ374𝑃7=284β‹…1.047β‰ˆ374The student population will be about 374 in 2020.

TRY IT #7

A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.

  1. ⓐWrite a formula for the number of hits.
  2. β“‘Estimate the number of hits in 5 weeks.

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