Learn on PengiOpenStax Algebra and TrigonometryChapter 13: Sequences, Probability, and Counting Theory

Lesson 13.3: Geometric Sequences

In this Grade 7 math lesson from OpenStax Algebra and Trigonometry, students learn to identify and work with geometric sequences by finding the common ratio, listing terms, and applying both recursive and explicit formulas. The lesson covers how to determine whether a sequence is geometric by dividing consecutive terms, and how to generate new terms by multiplying by a constant factor. Real-world contexts like annual salary growth are used to connect the concept of geometric sequences to exponential change.

Section 1

πŸ“˜ Geometric Sequences

New Concept

A geometric sequence is a list of numbers where each term is found by multiplying the previous one by a fixed number, the common ratio. We will explore how to find this ratio and use formulas to describe these sequences.

What’s next

Next, you'll master finding common ratios and terms through interactive examples. Then, you will apply recursive and explicit formulas in practice problems to solidify your understanding.

Section 2

Geometric Sequence and Common Ratio

Property

A 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 a1a_1 is the initial term of a geometric sequence and rr is the common ratio, the sequence will be

{a1,a1r,a1r2,a1r3,…}{\{a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots\}}

To determine if a set of numbers represents a geometric sequence, divide each term by the previous term. If the quotients are the same, a common ratio exists.

Section 3

Writing Terms of Geometric Sequences

Property

The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. To find the first four terms given the first term and the common factor:

  1. Multiply the initial term, a1a_1, by the common ratio to find the next term, a2a_2.
  2. Repeat the process, using an=anβˆ’1β‹…ra_n = a_{n-1} \cdot r to find a3a_3 and then a4a_4, until all four terms have been identified.
  3. Write the terms separated by commas within brackets.

Examples

  • List the first four terms of the geometric sequence with a1=3a_1 = 3 and r=4r = 4. The terms are {3,3β‹…4,12β‹…4,48β‹…4}\{3, 3 \cdot 4, 12 \cdot 4, 48 \cdot 4\}, which gives {3,12,48,192}\{3, 12, 48, 192\}.
  • List the first four terms of the geometric sequence with a1=80a_1 = 80 and r=12r = \frac{1}{2}. The terms are {80,80β‹…12,40β‹…12,20β‹…12}\{80, 80 \cdot \frac{1}{2}, 40 \cdot \frac{1}{2}, 20 \cdot \frac{1}{2}\}, which gives {80,40,20,10}\{80, 40, 20, 10\}.

Section 4

Recursive Formula for Geometric Sequences

Property

The recursive formula for a geometric sequence with common ratio rr and first term a1a_1 is

an=ranβˆ’1,nβ‰₯2a_n = r a_{n-1}, \quad n \geq 2

To write this formula, you must state the initial term a1a_1, find the common ratio rr by dividing any term by its preceding term, and substitute rr into the formula.

Section 5

Explicit Formula for Geometric Sequences

Property

Because a geometric sequence is an exponential function whose domain is the set of positive integers, we can write explicit formulas that allow us to find particular terms. The nnth term of a geometric sequence is given by the explicit formula:

an=a1rnβˆ’1a_n = a_1 r^{n-1}

Examples

  • Write an explicit formula for the sequence {3,12,48,192,…}\{3, 12, 48, 192, \ldots\}. The first term is a1=3a_1=3 and the common ratio is r=123=4r=\frac{12}{3}=4. The formula is an=3β‹…4nβˆ’1a_n = 3 \cdot 4^{n-1}.

Section 6

Solving Application Problems with Geometric Sequences

Property

In real-world scenarios involving geometric sequences, we may need to use an initial term of a0a_0 instead of a1a_1. In these problems, we can alter the explicit formula slightly by using the following formula:

an=a0rna_n = a_0 r^n

Examples

  • A new car is purchased for 30,000 dollars and depreciates by 20% each year. Its value after nn years is modeled by Vn=30000β‹…(1βˆ’0.20)n=30000β‹…(0.80)nV_n = 30000 \cdot (1 - 0.20)^n = 30000 \cdot (0.80)^n.

Book overview

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Chapter 13: Sequences, Probability, and Counting Theory

  1. Lesson 1

    Lesson 13.1: Sequences and Their Notations

  2. Lesson 2

    Lesson 13.2: Arithmetic Sequences

  3. Lesson 3Current

    Lesson 13.3: Geometric Sequences

  4. Lesson 4

    Lesson 13.4: Series and Their Notations

  5. Lesson 5

    Lesson 13.5: Counting Principles

  6. Lesson 6

    Lesson 13.6: Binomial Theorem

  7. Lesson 7

    Lesson 13.7: Probability

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Geometric Sequences

New Concept

A geometric sequence is a list of numbers where each term is found by multiplying the previous one by a fixed number, the common ratio. We will explore how to find this ratio and use formulas to describe these sequences.

What’s next

Next, you'll master finding common ratios and terms through interactive examples. Then, you will apply recursive and explicit formulas in practice problems to solidify your understanding.

Section 2

Geometric Sequence and Common Ratio

Property

A 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 a1a_1 is the initial term of a geometric sequence and rr is the common ratio, the sequence will be

{a1,a1r,a1r2,a1r3,…}{\{a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots\}}

To determine if a set of numbers represents a geometric sequence, divide each term by the previous term. If the quotients are the same, a common ratio exists.

Section 3

Writing Terms of Geometric Sequences

Property

The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. To find the first four terms given the first term and the common factor:

  1. Multiply the initial term, a1a_1, by the common ratio to find the next term, a2a_2.
  2. Repeat the process, using an=anβˆ’1β‹…ra_n = a_{n-1} \cdot r to find a3a_3 and then a4a_4, until all four terms have been identified.
  3. Write the terms separated by commas within brackets.

Examples

  • List the first four terms of the geometric sequence with a1=3a_1 = 3 and r=4r = 4. The terms are {3,3β‹…4,12β‹…4,48β‹…4}\{3, 3 \cdot 4, 12 \cdot 4, 48 \cdot 4\}, which gives {3,12,48,192}\{3, 12, 48, 192\}.
  • List the first four terms of the geometric sequence with a1=80a_1 = 80 and r=12r = \frac{1}{2}. The terms are {80,80β‹…12,40β‹…12,20β‹…12}\{80, 80 \cdot \frac{1}{2}, 40 \cdot \frac{1}{2}, 20 \cdot \frac{1}{2}\}, which gives {80,40,20,10}\{80, 40, 20, 10\}.

Section 4

Recursive Formula for Geometric Sequences

Property

The recursive formula for a geometric sequence with common ratio rr and first term a1a_1 is

an=ranβˆ’1,nβ‰₯2a_n = r a_{n-1}, \quad n \geq 2

To write this formula, you must state the initial term a1a_1, find the common ratio rr by dividing any term by its preceding term, and substitute rr into the formula.

Section 5

Explicit Formula for Geometric Sequences

Property

Because a geometric sequence is an exponential function whose domain is the set of positive integers, we can write explicit formulas that allow us to find particular terms. The nnth term of a geometric sequence is given by the explicit formula:

an=a1rnβˆ’1a_n = a_1 r^{n-1}

Examples

  • Write an explicit formula for the sequence {3,12,48,192,…}\{3, 12, 48, 192, \ldots\}. The first term is a1=3a_1=3 and the common ratio is r=123=4r=\frac{12}{3}=4. The formula is an=3β‹…4nβˆ’1a_n = 3 \cdot 4^{n-1}.

Section 6

Solving Application Problems with Geometric Sequences

Property

In real-world scenarios involving geometric sequences, we may need to use an initial term of a0a_0 instead of a1a_1. In these problems, we can alter the explicit formula slightly by using the following formula:

an=a0rna_n = a_0 r^n

Examples

  • A new car is purchased for 30,000 dollars and depreciates by 20% each year. Its value after nn years is modeled by Vn=30000β‹…(1βˆ’0.20)n=30000β‹…(0.80)nV_n = 30000 \cdot (1 - 0.20)^n = 30000 \cdot (0.80)^n.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 13: Sequences, Probability, and Counting Theory

  1. Lesson 1

    Lesson 13.1: Sequences and Their Notations

  2. Lesson 2

    Lesson 13.2: Arithmetic Sequences

  3. Lesson 3Current

    Lesson 13.3: Geometric Sequences

  4. Lesson 4

    Lesson 13.4: Series and Their Notations

  5. Lesson 5

    Lesson 13.5: Counting Principles

  6. Lesson 6

    Lesson 13.6: Binomial Theorem

  7. Lesson 7

    Lesson 13.7: Probability