Math

Comprehensive Study Guide to MYP5 Mathematics: Sequences, Functions, Equations, and Contextual Modeling

Foundational Diagnostic Mathematical Skills

• Order of Operations (BIDMAS / PEMDAS):

  • Standard operational hierarchy requires evaluating expressions in a strict sequence: Brackets / Parentheses ($\text{B/P}$) $\rightarrow$ Indices / Exponents ($\text{I/E}$) $\rightarrow$ Division and Multiplication from left to right ($\text{D/M}$) $\rightarrow$ Addition and Subtraction from left to right ($\text{A/S}$).

  • Canonical example: $2 + 3 \times 4 = 2 + 12 = 14$.

  • Worked diagnostic problems:

  • $5 + 2 \times (8 - 3) = 5 + 2 \times 5 = 5 + 10 = 15$

  • $(6 + 4) \times 3 - 7 = 10 \times 3 - 7 = 30 - 7 = 23$

  • $20 - 3^2 + 4 = 20 - 9 + 4 = 11 + 4 = 15$

  • $4 \times (5 - 2)^2 \div 6 = 4 \times 3^2 \div 6 = 4 \times 9 \div 6 = 36 \div 6 = 6$

• Fraction Operations and Simplification:

  • Adding or subtracting fractions requires determining a common denominator, converting numerators proportionally, combining numerators over the common denominator, and reducing to lowest terms.

  • Canonical example: $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$.

  • Worked diagnostic problems:

  • $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

  • $\frac{3}{5} + \frac{1}{4} = \frac{12}{20} + \frac{5}{20} = \frac{17}{20}$

  • $\frac{2}{3} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

  • $\frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}$

• Percentage Computations:

  • To calculate a percentage of an amount, convert the percentage to a decimal by dividing by $100$, then multiply by the given quantity.

  • Canonical example: $20\% \text{ of } 150 = 0.2 \times 150 = 30$.

  • Worked diagnostic problems:

  • $25\% \text{ of } 160 = 0.25 \times 160 = 40$

  • $60\% \text{ of } 45 = 0.60 \times 45 = 27$

  • $15\% \text{ of } 300 = 0.15 \times 300 = 45$

  • $8\% \text{ of } 250 = 0.08 \times 250 = 20$

• Scientific Notation (Standard Form):

  • Standard form is expressed as $a \times 10^n$, where the coefficient satisfies the strict inequality $1 \le a < 10$ and $n \in \mathbb{Z}$.

  • Canonical examples: $5300 = 5.3 \times 10^3$ and $0.0021 = 2.1 \times 10^{-3}$.

  • Worked diagnostic problems:

  • $62\,000 = 6.2 \times 10^4$

  • $0.00081 = 8.1 \times 10^{-4}$

  • $470\,000 = 4.7 \times 10^5$

  • $0.000036 = 3.6 \times 10^{-5}$

• Algebraic Manipulation:

  • Collecting Like Terms: Combine terms sharing identical variable components while preserving individual signs.

  • Canonical example: $4x + 3y - 2x + 5y = 2x + 8y$.

  • $6a + 2b - 3a + 4b = 3a + 6b$

  • $7x - 5x + 3y - y = 2x + 2y$

  • $5m + 2n - 3m - 6n = 2m - 4n$

  • $-3p + 4q + 7p - 2q = 4p + 2q$

  • Expanding Brackets: Apply the distributive property $a(b + c) = ab + ac$ across all interior terms, taking special care with negative signs.

  • Canonical example: $2(x + 3) = 2x + 6$.

  • $4(x - 2) + 3 = 4x - 8 + 3 = 4x - 5$

  • $3(2x + 1) - 5 = 6x + 3 - 5 = 6x - 2$

  • $2(3x - 4) - (x + 2) = 6x - 8 - x - 2 = 5x - 10$

  • $5(x + 2) - 3(x - 1) = 5x + 10 - 3x + 3 = 2x + 13$

  • Solving Linear Equations: Execute inverse operations symmetrically on both sides of the equality until the variable is isolated.

  • Canonical example: $3x + 5 = 20 \implies 3x = 15 \implies x = 5$.

  • $2x + 3 = 11 \implies 2x = 8 \implies x = 4$

  • $5x - 3 = 22 \implies 5x = 25 \implies x = 5$

  • $3(x + 2) = 18 \implies 3x + 6 = 18 \implies 3x = 12 \implies x = 4$

  • $4x + 7 = 2x + 15 \implies 2x = 8 \implies x = 4$

  • Factoring Quadratic Trinomials: For expressions of the form $x^2 + bx + c$, find two integers that multiply to $c$ and sum to $b$.

  • Canonical example: $x^2 + 5x + 6 = (x + 2)(x + 3)$.

  • $x^2 + 6x + 8 = (x + 2)(x + 4)$

  • $x^2 + 3x - 10 = (x + 5)(x - 2)$

  • $x^2 - 7x + 12 = (x - 3)(x - 4)$

  • $x^2 + 2x - 15 = (x + 5)(x - 3)$

  • Algebraic Substitution and Evaluation: Substitute numerical values inside parentheses in place of variables, then apply BIDMAS.

  • Canonical example: If $x = 2$, then $3x + 4 = 3(2) + 4 = 10$.

  • Given $a = 4$ and $b = -3$:

    • $2a + b = 2(4) + (-3) = 8 - 3 = 5$

    • $a^2 - b = (4)^2 - (-3) = 16 + 3 = 19$

    • $3a - 2b = 3(4) - 2(-3) = 12 + 6 = 18$

    • $a^2 + b^2 = (4)^2 + (-3)^2 = 16 + 9 = 25$

• Geometry and Trigonometry:

  • Pythagorean Theorem: In any right-angled triangle with legs $a$ and $b$ and hypotenuse $c$, $a^2 + b^2 = c^2$.

  • Canonical example: Legs $3\,cm$ and $4\,cm \implies c = \sqrt{3^2 + 4^2} = 5\,cm$.

  • Legs $5\,cm, 12\,cm \implies c = \sqrt{25 + 144} = 13\,cm$

  • Legs $9\,cm, 12\,cm \implies c = \sqrt{81 + 144} = \sqrt{225} = 15\,cm$

  • Hypotenuse $13\,cm$, leg $5\,cm \implies \text{other leg} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\,cm$

  • Legs $7\,cm, 24\,cm \implies c = \sqrt{49 + 576} = \sqrt{625} = 25\,cm$

  • Right-Triangle Trigonometric Ratios (SOH CAH TOA):

  • $\sin(\theta) = \frac{\text{opp}}{\text{hyp}}$, $\cos(\theta) = \frac{\text{adj}}{\text{hyp}}$, $\tan(\theta) = \frac{\text{opp}}{\text{adj}}$.

  • Canonical example: $\text{opp} = 3, \text{hyp} = 5 \implies \sin(\theta) = 0.6 \implies \theta \approx 36.9^\circ$.

  • $\text{opp} = 8, \text{hyp} = 17 \implies \sin(\theta) = \frac{8}{17} \implies \theta \approx 28.1^\circ$

  • $\text{adj} = 9, \text{hyp} = 15 \implies \cos(\theta) = \frac{9}{15} = 0.6 \implies \theta \approx 53.1^\circ$

  • $\text{opp} = 6, \text{adj} = 8 \implies \tan(\theta) = \frac{6}{8} = 0.75 \implies \theta \approx 36.9^\circ$

  • $\text{hyp} = 10, \theta = 30^\circ \implies \text{opp} = 10 \times \sin(30^\circ) = 10 \times 0.5 = 5$

  • Area and Perimeter Formulas:

  • Rectangle: $A = l \times w$, $P = 2(l + w)$.

  • Triangle: $A = \frac{1}{2}bh$.

  • Circle: $A = \pi r^2$, Circumference $C = 2\pi r$.

  • Square: $A = s^2$, $P = 4s$.

  • Worked diagnostic problems:

    • Rectangle $8\,cm \times 5\,cm \implies A = 40\,cm^2, P = 2(8 + 5) = 26\,cm$

    • Triangle base $10\,cm$, height $6\,cm \implies A = \frac{1}{2}(10)(6) = 30\,cm^2$

    • Circle radius $7\,cm$ ($\pi \approx 3.14$) $\implies A = 3.14 \times 7^2 = 3.14 \times 49 = 153.86 \approx 153.9\,cm^2$

    • Square side $9\,cm \implies A = 9^2 = 81\,cm^2, P = 4(9) = 36\,cm$

  • Geometric Angle Theorems:

  • Interior angles of a triangle sum to $180^\circ$.

  • Angles on a straight line sum to $180^\circ$.

  • Angles radiating around a single geometric point sum to $360^\circ$.

  • Worked diagnostic problems:

    • Triangle angles $55^\circ, 60^\circ, ? \implies 180^\circ - (55^\circ + 60^\circ) = 65^\circ$

    • Supplementary line angles $115^\circ, ? \implies 180^\circ - 115^\circ = 65^\circ$

    • Angles around a point $90^\circ, 120^\circ, 100^\circ, ? \implies 360^\circ - (90^\circ + 120^\circ + 100^\circ) = 50^\circ$

    • Right triangle angles $90^\circ, 35^\circ, ? \implies 180^\circ - (90^\circ + 35^\circ) = 55^\circ$

• Probability and Statistics:

  • Arithmetic Mean: The quotient of the sum of all values divided by the total number of values: $\bar{x} = \frac{\sum x_i}{n}$.

  • Data set $3, 7, 5, 9 \implies \frac{3 + 7 + 5 + 9}{4} = \frac{24}{4} = 6$

  • Data set $12, 15, 9, 18, 6 \implies \frac{60}{5} = 12$

  • Data set $20, 22, 18, 24 \implies \frac{84}{4} = 21$

  • Data set $5, 5, 5, 5, 10 \implies \frac{30}{5} = 6$

  • Theoretical Probability: Probability of an event is $P(E) = \frac{\text{favourable outcomes}}{\text{total possible outcomes}}$.

  • Rolling a $5$ on a fair six-sided die: $P(5) = \frac{1}{6}$.

  • Drawing a green marble from a bag with $6$ blue and $4$ green marbles: $P(\text{green}) = \frac{4}{10} = \frac{2}{5}$.

  • Spinner with $8$ equal sectors ($3$ yellow): $P(\text{yellow}) = \frac{3}{8}$.

  • Drawing a King from a standard deck of $52$ cards ($4$ Kings): $P(\text{King}) = \frac{4}{52} = \frac{1}{13}$.

  • Tabular Data Analysis:

  • Data distribution of favourite subjects across $30$ students:

    • Math: $8$

    • Science: $6$

    • English: $5$

    • Art: $7$

    • PE: $4$

  • Art or PE count: $7 + 4 = 11$.

  • Least popular subject: PE ($4$ students).

  • Math preference over English: $8 - 5 = 3$ more students.

  • Fraction choosing Science: $\frac{6}{30} = \frac{1}{5}$.

• Linear Functions and Coordinate Geometry:

  • In slope-intercept form $y = mx + c$, $m$ represents the gradient (slope) and $c$ denotes the vertical $y$-intercept.

  • $y = 4x - 3 \implies m = 4, c = -3$

  • $y = -2x + 7 \implies m = -2, c = 7$

  • $y = \frac{1}{2}x + 1 \implies m = \frac{1}{2}, c = 1$

  • $y = -x - 6 \implies m = -1, c = -6$

  • Evaluating coordinates on a line $y = 2x + 3$.

  • When $x = 0 \implies y = 2(0) + 3 = 3$

  • When $x = 2 \implies y = 2(2) + 3 = 7$

  • When $x = -1 \implies y = 2(-1) + 3 = 1$

  • When $x = 5 \implies y = 2(5) + 3 = 13$

  • Tabulating coordinates for linear functions across domain $x \in \{0, 1, 2, 3\}$:

  • $y = x + 2 \implies y \in \{2, 3, 4, 5\}$

  • $y = 2x - 1 \implies y \in \{-1, 1, 3, 5\}$

  • $y = -x + 4 \implies y \in \{4, 3, 2, 1\}$

  • $y = 3x \implies y \in \{0, 3, 6, 9\}$

  • Calculating gradient between coordinate pairs via $m = \frac{y_2 - y_1}{x_2 - x_1}$:

  • $(0, 1)$ and $(2, 5) \implies m = \frac{5 - 1}{2 - 0} = \frac{4}{2} = 2$

  • $(1, 4)$ and $(3, 10) \implies m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$

  • $(2, 8)$ and $(5, 2) \implies m = \frac{2 - 8}{5 - 2} = \frac{-6}{3} = -2$

  • $(0, 0)$ and $(4, 12) \implies m = \frac{12 - 0}{4 - 0} = \frac{12}{4} = 3$

Systems of Linear and Nonlinear Equations

• Solving Systems by Graphing:

  • A system consisting of a quadratic equation (parabola) and a linear equation (straight line) can yield zero, one, or two real solutions corresponding to the geometric points of intersection.

  • Step-by-step graphical method:

  • Step 1: Graph both equations accurately on the same Cartesian coordinate plane.

  • Step 2: Visually identify and estimate the coordinates of the intersection point(s).

  • Step 3: Verify algebraically that the estimated coordinate pair satisfies both original equations simultaneously.

  • Worked quadratic-linear example:

  • Equation 1: $y = x^2 - 2x - 1$

  • Equation 2: $y = -2x - 1$

  • Graphing both reveals an intersection at the point $(0, -1)$.

  • Verification in Equation 1: $-1 = (0)^2 - 2(0) - 1 \implies -1 = -1$ (True).

  • Verification in Equation 2: $-1 = -2(0) - 1 \implies -1 = -1$ (True).

  • Conclusion: The system has exactly one solution at $(0, -1)$.

• Solving Systems by Substitution:

  • The substitution method algebraically isolates one variable in one equation and substitutes the resulting expression into the second equation, reducing a two-variable system to a single-variable polynomial.

  • Worked quadratic-linear example:

  • Equation 1: $x^2 + x - y = -1$

  • Equation 2: $x + y = 4$

  • Step 1: Rearrange Equation 2 to isolate $y$: $y = -x + 4$.

  • Step 2: Substitute $(-x + 4)$ for $y$ into Equation 1:       $x^2 + x - (-x + 4) = -1$       $x^2 + x + x - 4 = -1$       $x^2 + 2x - 4 = -1$       $x^2 + 2x - 3 = 0$

  • Factor the quadratic expression in standard form:       $(x + 3)(x - 1) = 0$

  • Apply the Zero-Product Property:       $x + 3 = 0 \implies x = -3$       $x - 1 = 0 \implies x = 1$

  • Step 3: Back-substitute each $x$-value into $y = -x + 4$ to compute corresponding $y$-values:

    • For $x = -3$: $y = -(-3) + 4 = 3 + 4 = 7 \implies (-3, 7)$

    • For $x = 1$: $y = -(1) + 4 = 3 \implies (1, 3)$

  • Algebraic and graphical check confirms that the parabola $x^2 + x - y = -1$ and the line $x + y = 4$ intersect at exactly two points: $(-3, 7)$ and $(1, 3)$.

Arithmetic Sequences and Linear Modeling

• Fundamental Principles of Arithmetic Sequences:

  • An arithmetic sequence is an ordered list of numbers where the difference between consecutive terms is constant. This constant is designated as the common difference ($d$):     $d = a_{n} - a_{n-1}$

  • A sequence is arithmetic if and only if the difference between successive terms is invariant. A constant ratio does not constitute an arithmetic progression.

  • Core explicit formula for the $n$-th term:     $a_n = a_1 + (n - 1)d$     where $a_1$ represents the initial term, $d$ is the common difference, and $n \in \mathbb{Z}^+$ ($n \ge 1$) represents the term position.

  • Core recursive formulation:     $a_1 = \text{given value}, \quad a_n = a_{n-1} + d \quad (n \ge 2)$

  • Linear functional relationship:

  • An arithmetic sequence represents a discrete linear function whose domain is restricted to positive integers ($\mathbb{Z}^+$).

  • When plotted on a coordinate plane ($n$ on the horizontal axis, $a_n$ on the vertical axis), the terms lie along a straight line with a constant slope equal to the common difference $d$.

• Error Analysis and Common Misconceptions in Arithmetic Progressions:

  • Incorrect Explicit Formula Structure: A standard error involves omitting the shift $(n - 1)$ and writing $a_n = a_1 + nd$. In the sequence $14, 22, 30, 38, \dots$, where $a_1 = 14$ and $d = 8$, writing $a_n = 14 + 8n$ produces $a_1 = 22$, incorrectly skipping the starting term. The correct formula is $a_n = 14 + 8(n - 1) = 8n + 6$.

  • Sign Errors in Common Difference: In decreasing sequences like $2, 1, 0, -1, \dots$, stating $d = 1$ because each term is "1 less" is incorrect; the common difference must carry a negative sign ($d = -1$).

• Arithmetic Sequences Practice and Verification:

  • Sequence $19, 22, 25, 28, \dots$: $a_1 = 19, d = 3$. Next three terms: $31, 34, 37$.

  • Sequence $1, 12, 23, 34, \dots$: $a_1 = 1, d = 11$. Next three terms: $45, 56, 67$.

  • Sequence $60, 30, 0, -30, \dots$: $a_1 = 60, d = -30$. Next three terms: $-60, -90, -120$.

  • Sequence $\frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \frac{1}{3}, \dots$: Notice $\frac{5}{6}, \frac{4}{6}, \frac{3}{6}, \frac{2}{6}, \dots$ Common difference is $d = -\frac{1}{6}$. Next three terms: $\frac{1}{6}, 0, -\frac{1}{6}$.

  • Sequence $-5, -4, -3, -2, \dots$: Explicit equation is $a_n = -5 + 1(n - 1) = n - 6$. Tenth term: $a_{10} = 10 - 6 = 4$.

  • Sequence $-6, -9, -12, -15, \dots$: Explicit equation is $a_n = -6 - 3(n - 1) = -3n - 3$. Tenth term: $a_{10} = -3(10) - 3 = -33$.

  • Sequence $\frac{1}{2}, 1, 1\frac{1}{2}, 2, \dots$: Explicit equation is $a_n = \frac{1}{2} + \frac{1}{2}(n - 1) = \frac{1}{2}n$. Tenth term: $a_{10} = \frac{1}{2}(10) = 5$.

  • Sequence $\frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}, \dots$: Explicit equation is $a_n = \frac{3}{7} + \frac{1}{7}(n - 1) = \frac{1}{7}n + \frac{2}{7}$. Tenth term: $a_{10} = \frac{12}{7}$.

  • Geometric figure patterns:

  • Smiley face arrays: Group 1 has $4$ faces, Group 2 has $6$ faces, Group 3 has $8$ faces, Group 4 has $10$ faces. Sequence is $4, 6, 8, 10, \dots$ with common difference $d = 2$. This represents an arithmetic sequence.

  • Cube pyramids: Group 1 has $6$ cubes, Group 2 has $14$ cubes, Group 3 has $24$ cubes. Consecutive differences are $14 - 6 = 8$ and $24 - 14 = 10$. Because the differences are not constant, the sequence is not arithmetic.

• Nour's Karak Cart 5-Day Wall Challenge:

  • Context: Modeling savings and revenue growth for a mobile beverage cart.

  • Collaborative Team Roles: Sequence Tracker, Equation Builder, Function Connector, Prediction Checker, Wall Reporter.

  • Day 1 (Sequences as Functions - HSF.IF.A.3):

  • Starting equipment fund balance: $a(1) = 40\,SAR$.

  • Domain of function $a$: The set of positive integers $\mathbb{Z}^+ = \{1, 2, 3, 4, \dots\}$.

  • Domain restriction reasoning: The independent variable $n$ represents discrete elapsed business days; fractional days such as $n = 2.5$ cannot belong to the domain because financial accounting records are logged once at the end of each discrete business day.

  • Day 2 (The Recursive Rule - HSF.BF.A.2):

  • Daily fixed addition: $15\,SAR$.

  • Balance on Day 2: $a(2) = 55\,SAR$.

  • Recursive formulation: $a(1) = 40, \quad a(n) = a(n - 1) + 15 \quad \text{for } n \ge 2$.

  • Evaluating subsequent terms:

    • Day 3: $a(3) = a(2) + 15 = 55 + 15 = 70\,SAR$

    • Day 4: $a(4) = a(3) + 15 = 70 + 15 = 85\,SAR$

  • Day 3 (The Explicit Rule - HSF.BF.A.2):

  • Converting recursive form to explicit form:       $a(n) = a(1) + (n - 1)d = 40 + 15(n - 1) = 15n + 25$

  • Long-range direct prediction without repeated addition:       $a(30) = 15(30) + 25 = 450 + 25 = 475\,SAR$

  • Day 4 (Building a New Function - HSF.BF.A.1):

  • Cup sales function: $c(n) = 8 + 3(n - 1) = 3n + 5$, where $8$ cups are sold on Day 1 and volume grows by $3$ cups per day.

  • Unit profit: $2\,SAR$ profit per cup.

  • Profit function $P(n)$: $P(n) = 2 \cdot c(n) = 2[8 + 3(n - 1)] = 16 + 6(n - 1) = 6n + 10$.

  • Total money on hand $T(n)$ combining savings and daily sales profit:       $T(n) = a(n) + P(n) = (15n + 25) + (6n + 10) = 21n + 35$

  • Evaluation for Day 3:       $T(3) = 21(3) + 35 = 63 + 35 = 98\,SAR$       (Verification: $a(3) = 70\,SAR$, $c(3) = 8 + 3(2) = 14$ cups, $P(3) = 14 \times 2 = 28\,SAR$; $T(3) = 70 + 28 = 98\,SAR$).

  • Day 5 (Reading the Linear Model - HSF.LE.A.2):

  • Linear function from data points on the wall: $a(n) = 15n + 25$.

  • Practical interpretation of parameters:

    • Slope ($m = 15\,SAR/\text{day}$): The constant rate of daily savings added to the fund.

    • Theoretical $y$-intercept ($c = 25\,SAR$): The baseline value of the mathematical linear model when $n = 0$ (extrapolated value prior to Day 1).

  • Predicting Day 5 fund balance:       $a(5) = 15(5) + 25 = 75 + 25 = 100\,SAR$

• Arithmetic Sequences Mastery Pathway Protocol:

  • A 40-minute structured learning cycle focused on verified mastery over completion speed:

  • Station 1 (Emerging - "Build It"): 3 questions targeting pattern recognition, initial terms, and finding common differences.

  • Station 2 (Proficient - "Apply It"): 4 questions targeting the application of explicit and recursive models with confidence.

  • Station 3 (Advanced - "Challenge It"): 4 questions extending understanding to multi-step reasoning, contextual modeling, and algebraic proof.

  • Verification Mechanism: Progression requires passing a check by designated Station Verifiers, answering a quick verbal reasoning probe, rectifying all errors, and obtaining a formal stamp on the student's Math Passport.

  • Classroom Roles and Time Allocation:

  • Minutes $0 - 4$: Instructions, role assignments (Station Verifiers, Movement Monitor, Resource Manager).

  • Minutes $4 - 14$: Station 1 work.

  • Minutes $14 - 30$: Station progression under verification.

  • Minutes $30 - 36$: Consolidation and strengthening.

  • Minutes $36 - 40$: Exit reflection responding to three prompts: (1) Which station did I reach? (2) What skill is now secure? (3) What do I still need to strengthen?

Geometric Sequences and Exponential Modeling

• Fundamental Principles of Geometric Sequences:

  • A geometric sequence is a sequence of numbers where the ratio of consecutive terms is constant. This quotient is the common ratio ($r$):     $r = \frac{a_{n}}{a_{n-1}}$

  • Core explicit formula for the $n$-th term:     $a_n = a_1 r^{n - 1}$     where $a_1$ represents the initial term, $r$ is the common ratio, and the exponent is explicitly $n - 1$.

  • Core recursive formulation:     $a_1 = \text{given value}, \quad a_n = r \cdot a_{n-1} \quad (n \ge 2)$

  • Exponential functional relationship:

  • A geometric sequence with $r > 0$ and $r \ne 1$ represents an exponential function whose domain is restricted to a subset of integers.

• Discrete data points lie on a smooth continuous exponential curve $f(x) = a \cdot b^x$ because consecutive terms change by equal factors over equal intervals.

• Determining Sequence Types (Arithmetic vs. Geometric vs. Neither):

  • Test differences: If $a_2 - a_1 = a_3 - a_2 = \dots = d$, the sequence is arithmetic.

  • Test quotients: If $\frac{a_2}{a_1} = \frac{a_3}{a_2} = \dots = r$, the sequence is geometric.

  • Sequence $6, 12, 20, 30, 42, \dots$:     $\frac{12}{6} = 2, \quad \frac{20}{12} = \frac{5}{3}, \quad \frac{30}{20} = \frac{3}{2}, \quad \frac{42}{30} = \frac{7}{5}$     The ratios are not equal; thus, it is not geometric.

  • Sequence $256, 64, 16, 4, 1, \dots$:     $\frac{64}{256} = \frac{1}{4}, \quad \frac{16}{64} = \frac{1}{4}, \quad \frac{4}{16} = \frac{1}{4}, \quad \frac{1}{4} = \frac{1}{4}$     The common ratio is constant ($r = \frac{1}{4}$); therefore, it is geometric.

  • Sequence $27, 9, 3, 1, \frac{1}{3}, \dots$: Geometric ($r = \frac{1}{3}$).

  • Sequence $2, 6, 24, 120, 720, \dots$: Not geometric ($\frac{6}{2} = 3, \frac{24}{6} = 4, \frac{120}{24} = 5$; ratios are not constant).

  • Sequence $-1, 2, -4, 8, -16, \dots$: Geometric ($r = -2$).

  • Sequence $-7, -1, 5, 11, 17, \dots$: Arithmetic ($d = 6$).

  • Sequence $0, -1, -3, -7, -15, \dots$: Neither arithmetic nor geometric (differences are $-1, -2, -4, -8$; ratios are undefined or not constant).

  • Sequence $13.5, 40.5, 121.5, 364.5, \dots$: Geometric ($r = 3$).

• Deriving Explicit Equations for Geometric Sequences:

  • Given First Term and Common Ratio:

  • For $5, 15, 45, 135, \dots$: $a_1 = 5, r = 3 \implies a_n = 5(3)^{n-1}$.       Eighth term: $a_8 = 5(3)^{8 - 1} = 5(3)^7 = 5(2187) = 10\,935$.

  • For $88, -44, 22, -11, \dots$: $a_1 = 88, r = -\frac{1}{2} \implies a_n = 88\left(-\frac{1}{2}\right)^{n-1}$.       Eighth term: $a_8 = 88\left(-\frac{1}{2}\right)^7 = 88\left(-\frac{1}{128}\right) = -\frac{11}{16}$.

  • Given a Specific Term and Common Ratio:

  • Given $a_4 = 12$ and $r = 2$:

    • Substitute into general rule: $12 = a_1(2)^{4 - 1} \implies 12 = a_1(8) \implies a_1 = 1.5$.

    • Explicit rule: $a_n = 1.5(2)^{n-1}$.

    • Terms for $n \in \{1, 2, 3, 4, 5, 6\}$ are $1.5, 3, 6, 12, 24, 48$.

  • Given Two Arbitrary Terms:

  • Given $a_2 = 12$ and $a_5 = -768$:

    • Set up system: (1) $a_1 r = 12$, (2) $a_1 r^4 = -768$.

    • Divide equation (2) by (1): $\frac{a_1 r^4}{a_1 r} = \frac{-768}{12} \implies r^3 = -64 \implies r = -4$.

    • Solve for $a_1$: $a_1(-4) = 12 \implies a_1 = -3$.

    • Explicit rule: $a_n = -3(-4)^{n-1}$.

• Error Analysis in Geometric Sequences:

  • Misidentifying Exponent: If $a_2 = 48$ and $r = 6$, a common blunder is writing $a_n = a_1 r^n$, leading to $48 = a_1 (6)^2 \implies a_1 = \frac{48}{36} = \frac{4}{3}$. The true rule requires exponent $n - 1$, meaning $a_2 = a_1(6)^{2-1} = 6a_1 = 48 \implies a_1 = 8$, yielding $a_n = 8(6)^{n-1}$.

  • Inverting Base and Coefficient: Writing $a_n = r(a_1)^{n-1}$ flips the constant ratio and initial term, misrepresenting exponential compounding.

• Yara's Viral Video 5-Day Wall Challenge:

  • Context: Modeling viral video viewership and monetization using geometric growth.

  • Collaborative Team Roles: Growth Tracker, Ratio Detective, Formula Builder, Prediction Checker, Wall Reporter.

  • Day 1 (Sequences as Functions - HSF.IF.A.3):

  • Views on Day 1: $v(1) = 200$.

  • Domain of view function $v$: Set of positive integers $\mathbb{Z}^+ = \{1, 2, 3, 4, \dots\}$.

  • Non-integer domain argument: $n = 2.5$ cannot belong to the domain because viewing metrics are compiled and reported in daily discrete increments.

  • Day 2 (The Recursive Rule - HSF.BF.A.2):

  • Day 2 views: $v(2) = 400$.

  • Common ratio calculation: $r = \frac{v(2)}{v(1)} = \frac{400}{200} = 2$.

  • Recursive formulation: $v(1) = 200, \quad v(n) = 2 \cdot v(n - 1) \quad \text{for } n \ge 2$.

  • Projections: $v(3) = 2(400) = 800$ views; $v(4) = 2(800) = 1600$ views.

  • Day 3 (The Explicit Rule - HSF.BF.A.2):

  • Converting to explicit formula: $v(n) = v(1) \cdot r^{n - 1} = 200 \cdot 2^{n - 1}$.

  • Direct projection for Day 10:       $v(10) = 200 \cdot 2^{10 - 1} = 200 \cdot 2^9 = 200 \cdot 512 = 102\,400\text{ views}$

  • Day 4 (Total Views: Building a Series Function - HSA.SSE.B.4):

  • Total views over $n$ days represented by finite geometric series sum formula:       $S(n) = \frac{a(1 - r^n)}{1 - r}$

  • Evaluating for $n = 4$:       $S(4) = \frac{200(1 - 2^4)}{1 - 2} = \frac{200(1 - 16)}{-1} = \frac{200(-15)}{-1} = 3000\text{ total views}$

  • Manual summation check: $200 + 400 + 800 + 1600 = 3000$.

  • Building earnings function $E(n)$ at $0.05\,SAR$ per view:       $E(n) = 0.05 \cdot S(n)$       $E(4) = 0.05 \times 3000 = 150\,SAR$

  • Day 5 (Reading the Exponential Model - HSF.LE.A.2):

  • Expressing the model in standard exponential form: $f(n) = a \cdot b^n$.

  • Converting from $v(n) = 200(2)^{n-1} = 200(2^n \cdot 2^{-1}) = 100(2^n)$.

  • Parameters interpretation:

    • Base $b = 2$ represents the daily growth multiplier (views double every day).

    • Coefficient $a = 100$ represents the theoretical zero-day baseline ($n = 0$).

  • Evaluating Day 5 views: $f(5) = 100 \cdot 2^5 = 100 \times 32 = 3200\text{ views}$.

Arithmetic and Geometric Series

• Sum of a Finite Arithmetic Series:

  • An arithmetic series is the indicated sum of the terms of an arithmetic sequence.

  • General formulas for the sum of the first $n$ terms ($S_n$):     $S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}[2u_1 + (n - 1)d]$

• Sum of a Finite Geometric Series:

  • Algebraic derivation:     $S_n = a_1 + a_1r + a_1r^2 + \dots + a_1r^{n-1}$     Multiply by common ratio $r$:     $rS_n = a_1r + a_1r^2 + a_1r^3 + \dots + a_1r^n$     Subtract the two equations:     $S_n - rS_n = a_1 - a_1r^n$     $S_n(1 - r) = a_1(1 - r^n)$     Dividing by $(1 - r)$ (provided $r \ne 1$) yields:     $S_n = a_1 \left( \frac{1 - r^n}{1 - r} \right) = \frac{a_1(1 - r^n)}{1 - r}$

  • Worked examples:

  • Evaluate $\sum_{k=1}^{10} 4(3)^{k-1}$:       Here $a_1 = 4, r = 3, n = 10$.       $S_{10} = 4\left(\frac{1 - 3^{10}}{1 - 3}\right) = 4\left(\frac{1 - 59\,049}{-2}\right) = 4(29\,524) = 118\,096$

• Real-Life Financial Loan Amortization:

  • The formula for the monthly installment payment $M$ on a loan of amount $L$ with monthly interest rate $i$ (annual rate divided by 12) over $t$ months is:     $M = \frac{L}{\sum_{k=1}^t \left(\frac{1}{1 + i}\right)^k}$

  • The denominator represents a finite geometric series with first term $a_1 = \frac{1}{1 + i}$ and ratio $r = \frac{1}{1 + i}$.

  • Worked case study:

  • Loan amount $L = \$20\,000$, annual interest rate $6\% \implies i = \frac{0.06}{12} = 0.005$, loan term $5\text{ years} = 60\text{ months}$.

  • Sum of the geometric series in denominator:       $\sum_{k=1}^{60} \left(\frac{1}{1.005}\right)^k = \frac{\frac{1}{1.005}\left(1 - \left(\frac{1}{1.005}\right)^{60}\right)}{1 - \frac{1}{1.005}} \approx 51.72556$

  • Monthly payment computation:       $M = \frac{20\,000}{51.72556} \approx \$386.66$

  • Impact of rounding intermediate values: Rounding $\frac{1}{1.005} \approx 0.995$ introduces rounding error, generating an erroneous payment of $\$386.94$.

• Infinite Geometric Series and Convergence:

  • An infinite geometric series continues without end:     $\sum_{n=1}^\infty a_1 r^{n-1} = a_1 + a_1r + a_1r^2 + a_1r^3 + \dots$

  • The sequence of partial sums $S_n$ is defined by:     $S_n = a_1\left(\frac{1 - r^n}{1 - r}\right) = \frac{a_1}{1 - r} - \frac{a_1 r^n}{1 - r}$

  • Condition for Convergence:

  • If $|r| < 1$, as $n \to \infty$, $r^n \to 0$. Thus, the partial sums approach a finite limiting value:       $S = \frac{a_1}{1 - r}$

  • If $|r| \ge 1$, the sequence of partial sums does not approach a finite limit; the series diverges and has no sum.

• Applications of Infinite Geometric Series:

  • Damped Physical Oscillations (Pendulum Motion):

  • A pendulum travels $18\,\text{inches}$ on swing 1. Each subsequent swing travels $80\%$ of the preceding swing ($r = 0.8$).

  • Total distance traveled before coming to rest:       $S = \frac{a_1}{1 - r} = \frac{18}{1 - 0.8} = \frac{18}{0.2} = 90\,\text{inches} = 7.5\,\text{feet}$

  • If the first swing were $10\,\text{inches}$: $S = \frac{10}{1 - 0.8} = 50\,\text{inches}$.

  • Tire swing problem: First swing travels $14\,\text{feet}$, successive swings travel $75\%$ of previous ($r = 0.75$):       $S = \frac{14}{1 - 0.75} = \frac{14}{0.25} = 56\,\text{feet}$