MYP Inquiry Framework
Key Concept Patterns & Form
Related Concepts Generalization, Models, Change
Global Context Globalization and sustainability (Annuity schedules, compounding & fractals)
Statement of Inquiry Generalizing numerical and algebraic patterns allows the modeling of long-term change and the summation of infinite processes.
8.1

Arithmetic Progressions (AP): Terms & Series

  • Definition: A sequence where each term differs from the preceding term by a constant common difference ($d$): $$u_n - u_{n-1} = d$$
  • General ($n^{\text{th}}$) Term Formula: $$\mathbf{u_n = u_1 + (n - 1)d}$$
  • Arithmetic Series ($S_n$): The sum of the first $n$ terms. $$\mathbf{S_n = \frac{n}{2}\left(2u_1 + (n - 1)d\right) = \frac{n}{2}(u_1 + u_n)}$$
  • Proof (Gauss Method): Write the sum forwards and backwards: $$\begin{aligned} S_n &= u_1 + (u_1+d) + \dots + u_n \\ S_n &= u_n + (u_n-d) + \dots + u_1 \end{aligned}$$ Adding vertically gives $2S_n = n(u_1 + u_n) \implies S_n = \frac{n}{2}(u_1 + u_n)$.
Pascal's Triangle: Coefficients for (a + b)ⁿ 1 n = 0: (a+b)⁰ = 1 1 1 n = 1: 1a + 1b 1 2 1 n = 2: a² + 2ab + b² 1 3 3 1 n = 3: a³ + 3a²b + 3ab² + b³ 1 4 6 4 1 Each entry is the sum of the two numbers directly above it: 1 + 3 = 4, 3 + 3 = 6
8.2

Geometric Progressions (GP) & Infinite Series

  • Definition: A sequence where each term is obtained by multiplying the previous term by a non-zero common ratio ($r$): $$\frac{u_n}{u_{n-1}} = r \quad (r \ne 0)$$
  • General ($n^{\text{th}}$) Term Formula: $$\mathbf{u_n = u_1 \cdot r^{n - 1}}$$
  • Finite Geometric Series Sum ($S_n$): $$\mathbf{S_n = \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1(r^n - 1)}{r - 1}} \quad (r \ne 1)$$
  • Infinite Geometric Series Convergence ($S_\infty$): If and only if the absolute common ratio is strictly less than one ($|r| < 1$, i.e., $-1 < r < 1$), then as $n \to \infty$, $r^n \to 0$. The series converges to a finite sum: $$\mathbf{S_\infty = \frac{u_1}{1 - r}} \quad (|r| < 1)$$ If $|r| \ge 1$, the series diverges and has no finite sum.
8.3

Sigma Notation ($\sum$) & Summation Algebra

  • Notation Anatomy: $$\sum_{k=1}^n u_k = u_1 + u_2 + u_3 + \dots + u_n$$ Where $k$ is the dummy index of summation, $1$ is the lower limit, and $n$ is the upper limit.
  • Linearity Properties of Summation:
    • Constant Multiple Rule: $\sum_{k=1}^n c \cdot u_k = c \sum_{k=1}^n u_k$
    • Sum/Difference Rule: $\sum_{k=1}^n (u_k \pm v_k) = \sum_{k=1}^n u_k \pm \sum_{k=1}^n v_k$
    • Sum of a Constant: $\sum_{k=1}^n c = c + c + \dots + c = n \cdot c$
8.4

Pascal's Triangle & The Binomial Theorem

  • Binomial Coefficients $\binom{n}{r}$: $$\binom{n}{r} = \frac{n!}{r!(n - r)!}$$ Represents the $(r+1)^{\text{th}}$ entry in row $n$ of Pascal's triangle (where $r = 0, 1, 2, \dots, n$).
  • The Binomial Theorem: $$(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n - r} b^r = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \dots + \binom{n}{n} b^n$$
  • General Term Formula ($T_{r+1}$): $$\mathbf{T_{r+1} = \binom{n}{r} a^{n - r} b^r}$$ To find the term containing $x^k$ in $(Ax^p + B)^n$, equate the variable's net power to $k$ and solve for integer $r$.

Worked Examples (Criterion A & C)

Example 1 Finding Terms in an Arithmetic Progression

Problem: In an arithmetic sequence, the $4^{\text{th}}$ term is $19$ and the $11^{\text{th}}$ term is $54$.
(a) Find the first term $u_1$ and the common difference $d$.
(b) Find the sum of the first $20$ terms ($S_{20}$).

Step 1: Set up simultaneous equations using $u_n = u_1 + (n - 1)d$ $$u_4 = u_1 + 3d = 19 \quad \text{(Eq 1)}$$ $$u_{11} = u_1 + 10d = 54 \quad \text{(Eq 2)}$$
Step 2: Subtract equation (1) from equation (2) $$(u_1 - u_1) + (10d - 3d) = 54 - 19 \implies 7d = 35 \implies \mathbf{d = 5}$$
Step 3: Solve for $u_1$ $$u_1 + 3(5) = 19 \implies u_1 + 15 = 19 \implies \mathbf{u_1 = 4}$$
Step 4: Compute $S_{20}$ using series formula $$S_{20} = \frac{20}{2}[2(4) + (20 - 1)(5)] = 10[8 + 19(5)] = 10[8 + 95] = 10(103) = \mathbf{1030}$$ Criterion A: Complete algebraic simultaneous elimination with verified series sum
Example 2 Specific Term in Binomial Expansion

Problem: Find the coefficient of $x^3$ in the expansion of $(2x - 3)^5$.

Step 1: Write down the general term formula $$T_{r+1} = \binom{n}{r} a^{n - r} b^r$$

Here $n = 5$, $a = 2x$, and $b = -3$.

$$T_{r+1} = \binom{5}{r} (2x)^{5 - r} (-3)^r = \binom{5}{r} 2^{5 - r} (-3)^r x^{5 - r}$$
Step 2: Equate the power of $x$ to $3$ $$5 - r = 3 \implies r = 2$$
Step 3: Substitute $r = 2$ and evaluate $$T_3 = \binom{5}{2} 2^{5 - 2} (-3)^2 x^3 = \binom{5}{2} 2^3 (-3)^2 x^3$$

Evaluate components: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$, $2^3 = 8$, $(-3)^2 = 9$.

$$T_3 = 10 \times 8 \times 9 \cdot x^3 = 720 x^3$$

Therefore, the coefficient of $x^3$ is $\mathbf{720}$.

Criterion C: Explicit binomial parameter identification including sign fidelity

MYP 4–5 Practice Exercises: Sequences & Series

Tier 1 (AP/GP skills), Tier 2 (Series reasoning), and Tier 3 (Proofs and binomial applications).

Question 1 Tier 1: Standard 3 Marks

An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 4$.
(a) Find the $15^{\text{th}}$ term ($u_{15}$).
(b) Which term of this sequence equals $103$?

View Full Method Markscheme
[M1] Part (a): $u_{15} = u_1 + 14d = 7 + 14(4) = 7 + 56 = \mathbf{63}$.
[M1] Part (b): Set $u_n = 103 \implies 7 + (n - 1)4 = 103$.
[A1] $(n - 1)4 = 96 \implies n - 1 = 24 \implies \mathbf{n = 25}$ ($25^{\text{th}}$ term).
Question 2 Tier 1: Standard 3 Marks

Find the $8^{\text{th}}$ term of the geometric sequence: $3, -6, 12, -24, \dots$

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[M1] Find common ratio: $r = \frac{-6}{3} = -2$, and $u_1 = 3$.
[M1] Apply $u_n = u_1 \cdot r^{n-1}$: $u_8 = 3(-2)^{8 - 1} = 3(-2)^7$.
[A1] Calculate $(-2)^7 = -128$: $u_8 = 3 \times (-128) = \mathbf{-384}$.
Question 3 Tier 1: Standard 3 Marks

Evaluate the finite summation using arithmetic series formulas: $\sum_{k=1}^{30} (3k + 2)$.

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[M1] Identify first term $u_1 = 3(1) + 2 = 5$ and last term $u_{30} = 3(30) + 2 = 92$.
[M1] Number of terms is $n = 30$. Use $S_n = \frac{n}{2}(u_1 + u_n) = \frac{30}{2}(5 + 92)$.
[A1] Compute: $15 \times 97 = \mathbf{1455}$.
Question 4 Tier 2: Intermediate 4 Marks

A geometric sequence has first term $u_1 = 18$ and sum to infinity $S_\infty = 24$.
(a) Find the common ratio $r$.
(b) Calculate the third term $u_3$.

View Full Method Markscheme
[M1] Apply sum to infinity formula: $S_\infty = \frac{u_1}{1 - r} \implies 24 = \frac{18}{1 - r}$.
[M1] Solve for $r$: $24(1 - r) = 18 \implies 1 - r = \frac{18}{24} = \frac{3}{4} \implies \mathbf{r = \frac{1}{4} = 0.25}$.
[M1] Part (b): $u_3 = u_1 \cdot r^2 = 18\left(\frac{1}{4}\right)^2 = 18 \times \frac{1}{16}$.
[A1] Simplify: $\mathbf{u_3 = \frac{9}{8} = 1.125}$.
Question 5 Tier 2: Intermediate 4 Marks

Expand completely using Pascal's triangle: $(x - 2)^4$.

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[M1] Identify row $4$ coefficients from Pascal's triangle: $1, 4, 6, 4, 1$.
[M1] Write out expansion with terms $x$ and $(-2)$: $$1(x)^4 + 4(x)^3(-2)^1 + 6(x)^2(-2)^2 + 4(x)^1(-2)^3 + 1(-2)^4$$
[M1] Compute numerical coefficients: $$x^4 + 4(-2)x^3 + 6(4)x^2 + 4(-8)x + 16$$
[A1] Final expanded form: $\mathbf{x^4 - 8x^3 + 24x^2 - 32x + 16}$.
Question 6 Tier 2: Intermediate 4 Marks

Find the sum of all three-digit positive integers that are divisible by $7$.

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[M1] Smallest 3-digit multiple of $7$: $100 \div 7 = 14.28 \implies 15 \times 7 = 105$ ($u_1 = 105$).
[M1] Largest 3-digit multiple of $7$: $999 \div 7 = 142.71 \implies 142 \times 7 = 994$ ($u_n = 994$).
[M1] Number of terms: $994 = 105 + (n - 1)7 \implies 7(n - 1) = 889 \implies n - 1 = 127 \implies n = 128$.
[A1] Series sum: $S_{128} = \frac{128}{2}(105 + 994) = 64 \times 1099 = \mathbf{70336}$.
Question 7 Tier 2: Intermediate 4 Marks

Use the sum of an infinite geometric series to convert the recurring decimal $0.\overline{27} = 0.272727\dots$ into an irreducible vulgar fraction.

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[M1] Express as an infinite series: $0.\overline{27} = \frac{27}{100} + \frac{27}{10000} + \frac{27}{1000000} + \dots$
[M1] First term $u_1 = \frac{27}{100}$, common ratio $r = \frac{1}{100} = 0.01$ (note $|r| < 1$, so it converges).
[M1] Apply $S_\infty = \frac{u_1}{1 - r} = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99}$.
[A1] Reduce by dividing numerator and denominator by $9$: $\mathbf{\dfrac{3}{11}}$.
Question 8 Tier 3: Extended 5 Marks

In the expansion of $\left(x^2 + \dfrac{a}{x}\right)^6$, the constant term (independent of $x$) is $160$. Find all possible real values of $a$.

View Full Method Markscheme
[M1] General term: $T_{r+1} = \binom{6}{r} (x^2)^{6 - r} \left(\frac{a}{x}\right)^r = \binom{6}{r} x^{12 - 2r} \cdot a^r x^{-r} = \binom{6}{r} a^r x^{12 - 3r}$.
[M1] For the constant term, the power of $x$ must be zero: $12 - 3r = 0 \implies 3r = 12 \implies r = 4$.
[M1] Set term equal to $160$: $\binom{6}{4} a^4 = 160$.
[M1] Calculate $\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \implies 15 a^4 = 160 \implies a^4 = \frac{160}{15} = \frac{32}{3}$.
[A1] Solve for real $a$: $\mathbf{a = \pm \sqrt[4]{\frac{32}{3}} \approx \pm 1.808}$.
Question 9 Tier 3: Extended 5 Marks

Criterion B Investigation (Koch Snowflake Perimeter): An equilateral triangle has sides of length $1\text{ metre}$ (Step 0). At each subsequent step, the middle third of each side is replaced by two segments forming an outward-pointing equilateral triangle peak.
(a) Find the number of sides $N_k$ and the length of each side $L_k$ at step $k$.
(b) Write an expression for the total perimeter $P_k$ at step $k$.
(c) Deduce what happens to the perimeter as $k \to \infty$.

View Full Method Markscheme
[M1] Part (a): At step $0$, $N_0 = 3$ sides, each of length $L_0 = 1$. At each step, each segment is replaced by $4$ segments of length $\frac{1}{3}$. So $N_k = 3 \times 4^k$, and $L_k = \left(\frac{1}{3}\right)^k$.
[M1] Part (b): Perimeter $P_k = N_k \times L_k = 3 \times 4^k \times \left(\frac{1}{3}\right)^k = 3 \times \left(\frac{4}{3}\right)^k$.
[A1] This forms a geometric sequence with initial term $3$ and common ratio $r = \frac{4}{3}$.
[R1] Part (c): Because the common ratio $r = \frac{4}{3} > 1$, as $k \to \infty$, $\left(\frac{4}{3}\right)^k \to \infty$. Therefore, the total perimeter diverges to $\mathbf{\infty}$ (infinite perimeter enclosing a finite area).
Question 10 Tier 3: Extended 5 Marks

The numbers $x - 1$, $x + 2$, and $3x$ are the first three consecutive terms of a geometric sequence.
(a) Find the two possible values of $x$.
(b) Determine whether either value of $x$ produces a convergent infinite series, justifying your answer.

View Full Method Markscheme
[M1] For consecutive terms of a geometric sequence, the common ratio must be equal: $\frac{x + 2}{x - 1} = \frac{3x}{x + 2}$.
[M1] Cross-multiply: $(x + 2)^2 = 3x(x - 1) \implies x^2 + 4x + 4 = 3x^2 - 3x$.
[M1] Rearrange to quadratic: $2x^2 - 7x - 4 = 0 \implies (2x + 1)(x - 4) = 0$. Solutions: $\mathbf{x = 4 \quad \text{or} \quad x = -0.5}$.
[M1] Check common ratios:
  • If $x = 4$: terms are $3, 6, 12 \implies r = 2$. Since $|r| \ge 1$, this series diverges.
  • If $x = -0.5$: terms are $-1.5, 1.5, -1.5 \implies r = -1$. Since $|r| \ge 1$, this series also diverges.
[A1] Solutions are $\mathbf{x = 4}$ and $\mathbf{x = -\frac{1}{2}}$. Since $|r| \ge 1$ for both values, $\mathbf{\text{neither value produces a convergent infinite series}}$.

Sequences & Series Formulas

  • Arithmetic $n^{\text{th}}$ Term: $u_n = u_1 + (n - 1)d$
  • Arithmetic Sum: $S_n = \dfrac{n}{2}(2u_1 + (n - 1)d) = \dfrac{n}{2}(u_1 + u_n)$
  • Geometric $n^{\text{th}}$ Term: $u_n = u_1 \cdot r^{n - 1}$
  • Finite Geometric Sum: $S_n = \dfrac{u_1(1 - r^n)}{1 - r} \quad (r \ne 1)$
  • Infinite Geometric Sum: $S_\infty = \dfrac{u_1}{1 - r} \quad (|r| < 1)$
  • Binomial Coefficient: $\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$
  • Binomial General Term: $T_{r+1} = \binom{n}{r} a^{n - r} b^r$

Top Common MYP Pitfalls

1. Exponent Index in Geometric Sequences

Writing $u_n = u_1 \cdot r^n$ instead of $u_1 \cdot r^{n - 1}$. The first term has power $r^0 = 1$!

2. Applying $S_\infty$ When $|r| \ge 1$

Using $S_\infty = \frac{u_1}{1 - r}$ when $r = 2$ or $r = -1$. The formula is ONLY valid when $|r| < 1$.

3. Missing the Negative in Binomial Expansion

In $(2x - 5)^4$, setting $b = 5$ instead of $-5$. Powers $(-5)^r$ alternate signs (+, -, +, -, +)!

4. Off-by-One in General Term Formula

For the $4^{\text{th}}$ term $T_4$, using $r = 4$ instead of $r = 3$. Remember $T_{r+1}$ means $r = \text{term number} - 1$.

Interactive Mini Test: Sequences & Series

Answer the questions to see your score and feedback.

Question 1 of 5 Arithmetic Series Sum

What is the sum of the first $15$ terms of the arithmetic progression $4, 9, 14, 19, \dots$?

Question 2 of 5 Sum to Infinity

What is the sum to infinity of the geometric series $16 + 8 + 4 + 2 + \dots$?

Question 3 of 5 Pascal's Triangle Row

Which row of numbers represents the binomial coefficients for $(a + b)^5$?

Question 4 of 5 Binomial Expansion Term

In the expansion of $(x + 2)^6$, what is the coefficient of $x^4$?

Question 5 of 5 Geometric Convergence Condition

For which of the following common ratios does the infinite geometric series converge?