Comprehensive Study Guide: Geometric Sequences and Applications

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    • All: Lord, we turn our life and will over to You. That we will cease to struggle alone, but instead allow You to lift us up on eagle’s wings.
    • Leader: Saint Michael, defender of the Church of God, take us under your care and protection. And that in all things may we love and serve God and others.
    • All: This we humbly pray. Amen.

Course Overview and Learning Objectives

  • Institutional Details:

    • Institution: Saint Michael's College of Laguna (Established 1975 — "Levelling Up Our Legacy").
    • Term & Module: Term 1 MET 2 (Part 3) — Patterns in Numbers.
    • Curriculum Unit: Grade 11 General Mathematics, Unit 3: Sequences and Series, Lesson 2: Geometric Sequences (Quipper Platform).

  • Learning Competencies:

    • Illustrate the attributes of arithmetic and geometric sequences.
    • Solve problems involving arithmetic and geometric sequences.

  • Learning Targets:

    • Illustrate geometric sequences using numerical and real-life contexts.
    • Identify the common ratio and general form of a geometric sequence.
    • Solve real-life word problems involving geometric sequences.

  • Essential / Key Questions:

    • How do geometric sequences model real-life situations involving consistent growth or decay, such as savings, investments, or population change?
    • In what ways can understanding the common ratio and the general formula of a geometric sequence help us make accurate predictions and solve problems efficiently?

Real-World Contexts and Warm-Up Exploration

  • Real-World Scenarios:

    • Mobile Data Usage: Mobile data load decreases each time a fixed percentage is used.
    • Bacterial Growth: Bacteria multiply steadily over time.
    • Characteristics: These situations involve values that increase or decrease consistently and predictably, creating patterns modeled by geometric sequences.
    • Applications: Analyzing geometric sequences helps solve problems across finance, science, technology, and daily life.

  • Warm-Up Exploration Questions:

    • Scenario 1: A phone battery is at 80%80\% and loses 10%10\% of its current charge every hour. (Evaluates whether the battery decreases by the same amount or by the same percentage).
    • Scenario 2: An initial amount of ₱100\text{₱}100 doubles every week due to a savings challenge. (Evaluates whether money grows by addition or multiplication).
    • Scenario 3: A video gets 100100 views, and the view count becomes 33 times larger every hour. (Identifies the mathematical operation taking place each hour).

Review of Arithmetic Sequences

  • Definition of Arithmetic Sequence:

    • An arithmetic sequence (or arithmetic progression) is a sequence of numbers in which each term after the first is obtained by adding a constant value dd, known as the common difference, to the preceding term.
    • Increasing Condition: If d>0d > 0, the sequence is increasing.
    • Decreasing Condition: If d<0d < 0, the sequence is decreasing.

  • General Formula for Arithmetic Sequences:

    • an=a1+(n−1)da_n = a_1 + (n - 1)d
    • ana_n represents the nn\text{-th} term of the sequence.
    • a1a_1 represents the first term of the sequence.
    • nn represents the number of terms / term position.
    • dd represents the common difference.
    • Formula Manipulation for dd: d=an−a1n−1d = \frac{a_n - a_1}{n - 1}

  • Arithmetic Sequence Sample Problem:

    • Problem: Determine the 2222\text{-nd} term of the sequence: 4,7,10,13,164, 7, 10, 13, 16.
    • Given Values:
    • a1=4a_1 = 4
    • n=22n = 22
    • d=7−4=3d = 7 - 4 = 3
    • Step-by-Step Calculation:
    • an=a1+(n−1)da_n = a_1 + (n - 1)d
    • a22=4+(22−1)(3)a_{22} = 4 + (22 - 1)(3)
    • a22=4+(21)(3)a_{22} = 4 + (21)(3)
    • a22=4+63a_{22} = 4 + 63
    • a22=67a_{22} = 67

Fundamentals of Geometric Sequences

  • Definition of Geometric Sequence:

    • A geometric sequence (or geometric progression) is a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by a fixed, non-zero constant called the common ratio rr.

  • Common Ratio Formula:

    • r=anan−1r = \frac{a_n}{a_{n-1}}
    • The common ratio rr is found by dividing any term by the term immediately preceding it.

  • General Form of a Geometric Sequence:

    • Extended sequence notation: a1,a1r,a1r2,a1r3,…a_1, a_1 r, a_1 r^2, a_1 r^3, \dots
    • Formula for the nn\text{-th} term (ana_n):
    • an=a1rn−1a_n = a_1 r^{n-1}
    • ana_n is the term in the nn\text{-th} position.
    • a1a_1 is the first term.
    • rr is the common ratio.
    • nn is the term number.

  • Classification Drill: Geometric or Not:

    • Sequence A: 3,9,27,81,…3, 9, 27, 81, \dots
    • Classification: Geometric sequence.
    • Common Ratio: r=93=3r = \frac{9}{3} = 3
    • Sequence B: 5,10,15,20,…5, 10, 15, 20, \dots
    • Classification: Not a geometric sequence (It is an arithmetic sequence with d=5d = 5).
    • Sequence C: 64,32,16,8,…64, 32, 16, 8, \dots
    • Classification: Geometric sequence.
    • Common Ratio: r=3264=12r = \frac{32}{64} = \frac{1}{2}
    • Sequence D: 1,4,9,16,…1, 4, 9, 16, \dots
    • Classification: Not a geometric sequence (It is a sequence of perfect squares n2n^2).
    • Sequence E: −2,6,−18,54,…-2, 6, -18, 54, \dots
    • Classification: Geometric sequence.
    • Common Ratio: r=6−2=−3r = \frac{6}{-2} = -3

  • Basic Examples of Finding Terms:

    • Example 1: Given 4,8,16,32,64,…4, 8, 16, 32, 64, \dots
    • Ratio calculation: 84=2\frac{8}{4} = 2, 168=2\frac{16}{8} = 2, 3216=2\frac{32}{16} = 2, 6432=2\frac{64}{32} = 2.
    • Common ratio r=2r = 2.
    • Example 2: Given a1=5a_1 = 5 and r=2r = 2, find a4a_4
    • a4=5×24−1a_4 = 5 \times 2^{4-1}
    • a4=5×23a_4 = 5 \times 2^3
    • a4=5×8=40a_4 = 5 \times 8 = 40
    • Example 3: Find the 66\text{-th} term of a geometric sequence where a1=3a_1 = 3 and r=4r = 4
    • a6=3×46−1a_6 = 3 \times 4^{6-1}
    • a6=3×45a_6 = 3 \times 4^5
    • a6=3×1024=3072a_6 = 3 \times 1024 = 3072

  • Sequence Completion Review Drill:

    • Drill A: 3, 6, \text{\\_\_\_}, 24, \text{\\_\_\_}
    • Common ratio r=63=2r = \frac{6}{3} = 2
    • Missing 3rd term: 6×2=126 \times 2 = 12
    • Missing 5th term: 24×2=4824 \times 2 = 48
    • Completed sequence: 3,6,12,24,483, 6, 12, 24, 48
    • Drill B: 81, 27, 9, \text{\\_\_\_}, 1
    • Common ratio r=2781=13r = \frac{27}{81} = \frac{1}{3}
    • Missing 4th term: 9×13=39 \times \frac{1}{3} = 3
    • Completed sequence: 81,27,9,3,181, 27, 9, 3, 1
    • Drill C: 2, \text{\\_\_\_}, 8, -16, 32
    • Common ratio r=−168=−2r = \frac{-16}{8} = -2
    • Missing 2nd term: 2×(−2)=−42 \times (-2) = -4
    • Completed sequence: 2,−4,8,−16,322, -4, 8, -16, 32

Manipulating the General Formula of Geometric Sequences

  • Solving for the First Term (a1a_1):

    • Formula: a1=anrn−1a_1 = \frac{a_n}{r^{n-1}}
    • Example 1: Given a4=81a_4 = 81 and r=3r = 3, find a1a_1
    • a1=8134−1a_1 = \frac{81}{3^{4-1}}
    • a1=8133a_1 = \frac{81}{3^3}
    • a1=8127=3a_1 = \frac{81}{27} = 3
    • Example 2: Given a6=1024a_6 = 1024 and r=2r = 2, find a1a_1
    • a1=102426−1a_1 = \frac{1024}{2^{6-1}}
    • a1=102425a_1 = \frac{1024}{2^5}
    • a1=102432=32a_1 = \frac{1024}{32} = 32

  • Solving for the Common Ratio (rr):

    • Formula: r=ana1n−1r = \sqrt[n-1]{\frac{a_n}{a_1}}
    • Example 1: Given a1=2a_1 = 2 and a4=54a_4 = 54, find rr
    • r=5424−1r = \sqrt[4-1]{\frac{54}{2}}
    • r=273=3r = \sqrt[3]{27} = 3
    • Example 2: Given a1=3a_1 = 3 and a4=648a_4 = 648, find rr
    • r=64834−1r = \sqrt[4-1]{\frac{648}{3}}
    • r=2163=6r = \sqrt[3]{216} = 6

  • Solving for the Term Position (nn):

    • Formula: rn−1=ana1  ⟹  n=ln⁡(ana1)ln⁡(r)+1r^{n-1} = \frac{a_n}{a_1} \implies n = \frac{\ln\left(\frac{a_n}{a_1}\right)}{\ln(r)} + 1
    • Example 1: Given a1=5a_1 = 5, an=320a_n = 320, and r=2r = 2, find nn
    • 320=5×2n−1320 = 5 \times 2^{n-1}
    • 64=2n−164 = 2^{n-1}
    • 26=2n−1  ⟹  n−1=6  ⟹  n=72^6 = 2^{n-1} \implies n - 1 = 6 \implies n = 7
    • Example 2: Given a1=3a_1 = 3, an=243a_n = 243, and r=3r = 3, find nn
    • 243=3×3n−1243 = 3 \times 3^{n-1}
    • 81=3n−181 = 3^{n-1}
    • 34=3n−1  ⟹  n−1=4  ⟹  n=53^4 = 3^{n-1} \implies n - 1 = 4 \implies n = 5
    • Example 3: Given a1=5a_1 = 5, an=1,310,720a_n = 1,310,720, and r=4r = 4, find nn
    • 1,310,720=5×4n−11,310,720 = 5 \times 4^{n-1}
    • 262,144=4n−1262,144 = 4^{n-1}
    • 49=4n−1  ⟹  n−1=9  ⟹  n=104^9 = 4^{n-1} \implies n - 1 = 9 \implies n = 10

Real-World Applications: Exponential Growth and Decay

  • Categorization of Applications:

    • Geometric sequences model real-life situations where quantities change at a constant multiplicative rate, divided into exponential growth and exponential decay.

  • 1. Exponential Growth:

    • Definition: Occurs when a quantity increases by a fixed percentage or multiplies by a constant factor over equal time intervals.
    • Condition: In geometric sequences, exponential growth occurs when r>1r > 1.
    • Real-World Example: A village has an initial population of 1,0001,000 people, and the population doubles each year. Calculate the population in the fifth year.
    • Given: a1=1,000a_1 = 1,000, r=2r = 2, n=5n = 5
    • Calculation:
      • a5=a1r5−1a_5 = a_1 r^{5-1}
      • a5=1000×24a_5 = 1000 \times 2^4
      • a5=1000×16=16,000a_5 = 1000 \times 16 = 16,000
    • Conclusion: There will be 16,00016,000\text{ people} in the fifth year.

  • 2. Exponential Decay:

    • Definition: Occurs when a quantity decreases by a constant fraction or percentage over equal time intervals.
    • Condition: In geometric sequences, exponential decay occurs when the common ratio rr satisfies 0<r<10 < r < 1
    • Real-World Example: A car's value depreciates by 20%20\% each year. If the original cost of the car is ₱800,000\text{₱}800,000, calculate its value at the end of the fifth year.
    • Given:
      • a1=800,000a_1 = 800,000 (initial value)
      • r=1−0.20=0.80r = 1 - 0.20 = 0.80 (retaining 80%80\% of its value yearly)
      • n=5n = 5
    • Calculation:
      • a5=a1r5−1a_5 = a_1 r^{5-1}
      • a5=800,000×(0.80)5−1a_5 = 800,000 \times (0.80)^{5-1}
      • a5=800,000×(0.80)4a_5 = 800,000 \times (0.80)^4
      • a5=800,000×0.4096a_5 = 800,000 \times 0.4096
      • a5=327,680a_5 = 327,680
    • Conclusion: The car's value at the end of the fifth year is ₱327,680\text{₱}327,680

Detailed Practice Problems

  • Practice Problem 1: General Term of a Sequence:

    • Given Sequence: 4,16,64,256,…4, 16, 64, 256, \dots
    • Part A: Determine the common ratio.
    • Formula: r=a2a1r = \frac{a_2}{a_1}
    • r=164=4r = \frac{16}{4} = 4
    • Part B: Determine the general formula for the nn\text{-th} term.
    • Formula: an=a1rn−1a_n = a_1 r^{n-1}
    • an=4×4n−1=4na_n = 4 \times 4^{n-1} = 4^n

  • Practice Problem 2: Thawing Organic Matter Decay:

    • Scenario: A piece of frozen organic matter weighs 7272\text{ grams}. It loses one-third of its remaining mass each day due to a thawing process.
    • Given Parameters:
    • a1=72a_1 = 72
    • r=1−13=23r = 1 - \frac{1}{3} = \frac{2}{3} (retains two-thirds of its mass daily)
    • Part A: What is the mass on the second day (n=2n = 2)?
    • a2=a1r2−1a_2 = a_1 r^{2-1}
    • a2=72×(23)1=48a_2 = 72 \times \left(\frac{2}{3}\right)^1 = 48\text{ grams}
    • Part B: What is the mass on the fifth day (n=5n = 5)?
    • a5=a1r5−1a_5 = a_1 r^{5-1}
    • a5=72×(23)4a_5 = 72 \times \left(\frac{2}{3}\right)^4
    • a5=72×1681a_5 = 72 \times \frac{16}{81}
    • a5=115281≈14.22a_5 = \frac{1152}{81} \approx 14.22\text{ grams}

Summary of Key Concepts

  • Core Definition: A geometric sequence is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio (rr).
  • Common Ratio: Determined by dividing any term by the preceding term: r=anan−1r = \frac{a_n}{a_{n-1}}.
  • General Term Formula: The nn\text{-th} term of a geometric sequence is calculated using an=a1rn−1a_n = a_1 r^{n-1}.

Attributions and References

  • Media Attributions:

    • Slide 3 & 10 Image: Live Bacteria in a Petri Dish by Edward Jenner (Licensed under Pexels License via Pexels).
    • Slide 11 Image: Photo Of Woman Looking Through Microscope by Artem Podrez (Licensed under Pexels License via Pexels).

  • References:

    • Pierce, Rod. "Geometric Sequences and Sums." Accessed June 19, 2025. https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
    • Math LibreTexts. "9.3: Geometric Sequences and Series." Last modified July 18, 2022. https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/09%3A_Sequences_Series_and_the_Binomial_Theorem/9.03%3A_Geometric_Sequences_and_Series
    • Alamo Colleges District. "Geometric Sequences." Accessed June 19, 2025. https://www.alamo.edu/contentassets/afe30946fa58450c89840c1173f3b9d0/sequences/math1314-geometric-sequences.pdf
    • Cuemath. "Geometric Sequence Formulas." Accessed June 19, 2025. https://www.cuemath.com/geometric-sequence-formulas/