Learn on PengiSaxon Algebra 1Chapter 11: Advanced Topics in Algebra

Lesson 105: Recognizing and Extending Geometric Sequences

In this Grade 9 Saxon Algebra 1 lesson from Chapter 11, students learn to identify geometric sequences by finding the common ratio between consecutive terms and apply the nth-term formula A(n) = ar^(n-1) to extend sequences and calculate specific terms. The lesson covers sequences with positive, negative, and fractional common ratios through worked examples involving integers and fractions.

Section 1

πŸ“˜ Recognizing and Extending Geometric Sequences

New Concept

A geometric sequence is a sequence with a constant ratio between consecutive terms.

What’s next

Next, you’ll practice finding the common ratio and use it to extend sequences and calculate any term with a powerful formula.

Section 2

Geometric sequence

Property

A geometric sequence is a sequence with a constant ratio between consecutive terms. The ratio between consecutive terms is known as the common ratio.

Explanation

Think of it as a number pattern where you multiply by the same secret number to get from one term to the next! This special multiplier, called the common ratio, is the key to the whole sequence. To find it, just divide any term by the one that came right before it. It’s a chain reaction of multiplication!

Examples

  • In the sequence 3,15,75,...3, 15, 75, ..., find the common ratio (153=5\frac{15}{3}=5) and multiply to find the next term: 75Γ—5=37575 \times 5 = 375.
  • For 100,βˆ’50,25,...100, -50, 25, ..., the ratio is βˆ’12-\frac{1}{2}. The next term is 25Γ—(βˆ’12)=βˆ’12.525 \times (-\frac{1}{2})= -12.5.
  • The sequence 2,8,32,128,...2, 8, 32, 128, ... has a common ratio of 4. The next two terms are 128Γ—4=512128 \times 4 = 512 and 512Γ—4=2048512 \times 4 = 2048.

Section 3

Example Card: Extending Geometric Sequences

Let's see how a single multiplier generates an entire sequence, even with alternating signs. This example shows how to extend a sequence, a key idea from this lesson.

Example Problem

Find the next four terms in the geometric sequence 3,βˆ’12,48,βˆ’192,...3, -12, 48, -192, ...

Section 4

Common ratio

Property

The ratio between consecutive terms is known as the common ratio. In a geometric sequence, the ratio of any term divided by the previous term is the same for any two consecutive terms.

Explanation

The common ratio is the secret code of a geometric sequence! It's the one number you're always multiplying by to continue the pattern. To crack the code, just pick any two neighbors in the sequence and divide the second one by the first one. The result is your constant multiplier, whether it's a whole number, a fraction, or negative.

Examples

  • In the sequence 4,12,36,108,...4, 12, 36, 108, ..., the common ratio is 124=3\frac{12}{4} = 3.
  • For 320,βˆ’80,20,βˆ’5,...320, -80, 20, -5, ..., the common ratio is βˆ’80320=βˆ’14\frac{-80}{320} = -\frac{1}{4}. The signs flip because the ratio is negative!
  • In 0.4,1,2.5,6.25,...0.4, 1, 2.5, 6.25, ..., the common ratio is 10.4=2.5\frac{1}{0.4} = 2.5.

Section 5

Finding the nth Term of a Geometric Sequence

Property

Let A(n)A(n) equal the nnth term of a geometric sequence, then

A(n)=arnβˆ’1A(n) = ar^{n-1}
where aa is the first term of the sequence and rr is the common ratio.

Explanation

Want to jump to any term in a sequence without listing them all? This formula is your mathematical time machine! Just plug in the first term (aa), the common ratio (rr), and the term number you want (nn). Remember the tricky part: the exponent is always one less than the term number (nβˆ’1n-1) because the first term doesn't get multiplied yet!

Examples

  • Find the 5th term if a=6a=6 and r=βˆ’2r=-2: A(5)=6(βˆ’2)5βˆ’1=6(βˆ’2)4=6(16)=96A(5) = 6(-2)^{5-1} = 6(-2)^4 = 6(16) = 96.
  • Find the 8th term of 12,14,18,...\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, .... Here a=12a=\frac{1}{2} and r=12r=\frac{1}{2}. A(8)=12(12)8βˆ’1=(12)8=1256A(8) = \frac{1}{2}(\frac{1}{2})^{8-1} = (\frac{1}{2})^8 = \frac{1}{256}.
  • A ball's first bounce is 1.5 yards high (a=1.5a=1.5), and each bounce is 80% of the last (r=0.8r=0.8). The 5th bounce height is A(5)=1.5(0.8)5βˆ’1β‰ˆ0.61A(5) = 1.5(0.8)^{5-1} \approx 0.61 yards.

Book overview

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Chapter 11: Advanced Topics in Algebra

  1. Lesson 1

    Lesson 101: Solving Multi-Step Absolute-Value Inequalities

  2. Lesson 2

    Lesson 102: Solving Quadratic Equations Using Square Roots

  3. Lesson 3

    Lesson 103: Dividing Radical Expressions

  4. Lesson 4

    Lesson 104: Solving Quadratic Equations by Completing the Square

  5. Lesson 5Current

    Lesson 105: Recognizing and Extending Geometric Sequences

  6. Lesson 6

    Lesson 106: Solving Radical Equations

  7. Lesson 7

    Lesson 107: Graphing Absolute-Value Functions

  8. Lesson 8

    Lesson 108: Identifying and Graphing Exponential Functions

  9. Lesson 9

    Lesson 109: Graphing Systems of Linear Inequalities

  10. Lesson 10

    Lesson 110: Using the Quadratic Formula

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Recognizing and Extending Geometric Sequences

New Concept

A geometric sequence is a sequence with a constant ratio between consecutive terms.

What’s next

Next, you’ll practice finding the common ratio and use it to extend sequences and calculate any term with a powerful formula.

Section 2

Geometric sequence

Property

A geometric sequence is a sequence with a constant ratio between consecutive terms. The ratio between consecutive terms is known as the common ratio.

Explanation

Think of it as a number pattern where you multiply by the same secret number to get from one term to the next! This special multiplier, called the common ratio, is the key to the whole sequence. To find it, just divide any term by the one that came right before it. It’s a chain reaction of multiplication!

Examples

  • In the sequence 3,15,75,...3, 15, 75, ..., find the common ratio (153=5\frac{15}{3}=5) and multiply to find the next term: 75Γ—5=37575 \times 5 = 375.
  • For 100,βˆ’50,25,...100, -50, 25, ..., the ratio is βˆ’12-\frac{1}{2}. The next term is 25Γ—(βˆ’12)=βˆ’12.525 \times (-\frac{1}{2})= -12.5.
  • The sequence 2,8,32,128,...2, 8, 32, 128, ... has a common ratio of 4. The next two terms are 128Γ—4=512128 \times 4 = 512 and 512Γ—4=2048512 \times 4 = 2048.

Section 3

Example Card: Extending Geometric Sequences

Let's see how a single multiplier generates an entire sequence, even with alternating signs. This example shows how to extend a sequence, a key idea from this lesson.

Example Problem

Find the next four terms in the geometric sequence 3,βˆ’12,48,βˆ’192,...3, -12, 48, -192, ...

Section 4

Common ratio

Property

The ratio between consecutive terms is known as the common ratio. In a geometric sequence, the ratio of any term divided by the previous term is the same for any two consecutive terms.

Explanation

The common ratio is the secret code of a geometric sequence! It's the one number you're always multiplying by to continue the pattern. To crack the code, just pick any two neighbors in the sequence and divide the second one by the first one. The result is your constant multiplier, whether it's a whole number, a fraction, or negative.

Examples

  • In the sequence 4,12,36,108,...4, 12, 36, 108, ..., the common ratio is 124=3\frac{12}{4} = 3.
  • For 320,βˆ’80,20,βˆ’5,...320, -80, 20, -5, ..., the common ratio is βˆ’80320=βˆ’14\frac{-80}{320} = -\frac{1}{4}. The signs flip because the ratio is negative!
  • In 0.4,1,2.5,6.25,...0.4, 1, 2.5, 6.25, ..., the common ratio is 10.4=2.5\frac{1}{0.4} = 2.5.

Section 5

Finding the nth Term of a Geometric Sequence

Property

Let A(n)A(n) equal the nnth term of a geometric sequence, then

A(n)=arnβˆ’1A(n) = ar^{n-1}
where aa is the first term of the sequence and rr is the common ratio.

Explanation

Want to jump to any term in a sequence without listing them all? This formula is your mathematical time machine! Just plug in the first term (aa), the common ratio (rr), and the term number you want (nn). Remember the tricky part: the exponent is always one less than the term number (nβˆ’1n-1) because the first term doesn't get multiplied yet!

Examples

  • Find the 5th term if a=6a=6 and r=βˆ’2r=-2: A(5)=6(βˆ’2)5βˆ’1=6(βˆ’2)4=6(16)=96A(5) = 6(-2)^{5-1} = 6(-2)^4 = 6(16) = 96.
  • Find the 8th term of 12,14,18,...\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, .... Here a=12a=\frac{1}{2} and r=12r=\frac{1}{2}. A(8)=12(12)8βˆ’1=(12)8=1256A(8) = \frac{1}{2}(\frac{1}{2})^{8-1} = (\frac{1}{2})^8 = \frac{1}{256}.
  • A ball's first bounce is 1.5 yards high (a=1.5a=1.5), and each bounce is 80% of the last (r=0.8r=0.8). The 5th bounce height is A(5)=1.5(0.8)5βˆ’1β‰ˆ0.61A(5) = 1.5(0.8)^{5-1} \approx 0.61 yards.

Book overview

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

Continue this chapter

Chapter 11: Advanced Topics in Algebra

  1. Lesson 1

    Lesson 101: Solving Multi-Step Absolute-Value Inequalities

  2. Lesson 2

    Lesson 102: Solving Quadratic Equations Using Square Roots

  3. Lesson 3

    Lesson 103: Dividing Radical Expressions

  4. Lesson 4

    Lesson 104: Solving Quadratic Equations by Completing the Square

  5. Lesson 5Current

    Lesson 105: Recognizing and Extending Geometric Sequences

  6. Lesson 6

    Lesson 106: Solving Radical Equations

  7. Lesson 7

    Lesson 107: Graphing Absolute-Value Functions

  8. Lesson 8

    Lesson 108: Identifying and Graphing Exponential Functions

  9. Lesson 9

    Lesson 109: Graphing Systems of Linear Inequalities

  10. Lesson 10

    Lesson 110: Using the Quadratic Formula