1.2

The Common Difference & General $n$th Term Formula

  • Definition of an Arithmetic Sequence: A sequence in which the difference between any two consecutive terms is a constant $d \in \mathbb{R}$: $$d = u_{k+1} - u_k = u_2 - u_1 = u_3 - u_2$$ If $d > 0$, the sequence is strictly increasing; if $d < 0$, the sequence is strictly decreasing.
  • The General $n$th Term Formula: $$u_n = u_1 + (n - 1)d$$ where $u_1$ is the first term, $d$ is the common difference, and $n \in \mathbb{Z}^+$ is the term position. Notice the $(n - 1)$ factor: to reach the $n$th term from the first term, you add the common difference $(n - 1)$ times.
  • Consecutive Terms Condition (The Arithmetic Mean): Three numbers $a, b, c$ form consecutive terms of an arithmetic progression if and only if: $$b - a = c - b \iff 2b = a + c \iff b = \frac{a + c}{2}$$

Example 1

Finding General Terms & Identifying Positions

An arithmetic sequence begins $107, 101, 95, 89, \dots$
(a) Find an explicit formula for the $n$th term $u_n$.
(b) Determine the position of the first negative term in the sequence.

  1. Part (a): Identify first term $u_1$ and common difference $d$: $$u_1 = 107, \quad d = 101 - 107 = -6$$ Substitute into the general term formula: $$u_n = 107 + (n - 1)(-6) = 107 - 6n + 6 = 113 - 6n$$
  2. Part (b): Set up the inequality for negative terms ($u_n < 0$): $$113 - 6n < 0 \implies 6n > 113 \implies n > \frac{113}{6} \approx 18.833$$
  3. State the smallest integer $n$: Since $n \in \mathbb{Z}^+$, the smallest integer satisfying $n > 18.833$ is $n = 19$.
    Checking: $u_{18} = 113 - 6(18) = 5 > 0$ and $u_{19} = 113 - 6(19) = -1 < 0$.

Answer: (a) $u_n = 113 - 6n$; (b) The $19$th term ($u_{19} = -1$)

1.2

Simultaneous Systems to Find $u_1$ and $d$

  • Problem Structure: Frequently, IB exam questions give two non-consecutive terms (e.g. $u_5$ and $u_{12}$) without stating $u_1$ or $d$.
  • Method of Subtraction: Write both equations in terms of $u_1$ and $d$: $$\begin{cases} u_p = u_1 + (p - 1)d \\ u_q = u_1 + (q - 1)d \end{cases}$$ Subtracting the first equation from the second eliminates $u_1$ immediately: $$u_q - u_p = (q - p)d \implies d = \frac{u_q - u_p}{q - p}$$
  • Back-Substitution: Once $d$ is determined, substitute it back into either original equation to calculate $u_1$.

Example 2

Solving an Arithmetic Sequence System

In an arithmetic sequence, the $4$th term is $19$ and the $11$th term is $54$. Find the first term $u_1$, the common difference $d$, and calculate $u_{25}$.

  1. Formulate equations for $u_4$ and $u_{11}$: $$u_4 = u_1 + 3d = 19 \quad \text{--- (1)}$$ $$u_{11} = u_1 + 10d = 54 \quad \text{--- (2)}$$
  2. Subtract (1) from (2) to eliminate $u_1$: $$(u_1 + 10d) - (u_1 + 3d) = 54 - 19$$ $$7d = 35 \implies d = 5$$
  3. Substitute $d = 5$ into equation (1) to find $u_1$: $$u_1 + 3(5) = 19 \implies u_1 + 15 = 19 \implies u_1 = 4$$
  4. Calculate $u_{25}$: $$u_{25} = u_1 + 24d = 4 + 24(5) = 4 + 120 = 124$$

Answer: $u_1 = 4$, $d = 5$, and $u_{25} = 124$

1.2

The Arithmetic Series & Gauss Summation Formulas

  • Arithmetic Series Definition: The sum of the first $n$ terms of an arithmetic sequence: $$S_n = u_1 + u_2 + u_3 + \dots + u_n$$
  • The Two Standard IB Formula Booklet Forms:
    • When the last term $u_n$ is known: $$S_n = \frac{n}{2}(u_1 + u_n)$$
    • When only $u_1$ and $d$ are known: Substitute $u_n = u_1 + (n - 1)d$ into the formula above: $$S_n = \frac{n}{2}\left(2u_1 + (n - 1)d\right)$$
  • Geometric Intuition (Gauss Inversion Trick): Writing the sum forwards and backwards pairs each term into a constant sum $(u_1 + u_n)$. Summing $n$ identical pairs gives $2S_n = n(u_1 + u_n)$, yielding the formula.
Arithmetic Series Step Progression and Gauss Pairing Illustrates discrete column heights increasing by common difference d, and pairing with an inverted series to form a constant-height rectangle. u₁ u₂ u₃ u₄ +d Sₙ = u₁ + u₂ + … + uₙ Gauss Inversion Principle Sₙ = u₁ + u₂ + … + uₙ Sₙ = uₙ + uₙ₋₁ + … + u₁ 2Sₙ = n(u₁ + uₙ) Sₙ = ⁿ⁄₂ (u₁ + uₙ)
Figure 1.2: Discrete step progression with constant difference $d$, and the Gauss inversion technique yielding $S_n = \frac{n}{2}(u_1 + u_n)$.

Example 3

Evaluating an Arithmetic Sum

An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 4$. Find the sum of the first $25$ terms.

  1. Identify given parameters: $$u_1 = 7, \quad d = 4, \quad n = 25$$
  2. Use the standard formula $S_n = \frac{n}{2}[2u_1 + (n - 1)d]$: $$S_{25} = \frac{25}{2}\left[ 2(7) + (25 - 1)(4) \right]$$
  3. Evaluate inside the brackets: $$2(7) + 24(4) = 14 + 96 = 110$$
  4. Compute the final total: $$S_{25} = \frac{25}{2}(110) = 25 \times 55 = 1375$$

Answer: $S_{25} = 1375$

1.2

Sigma Notation & Quadratic Solving for $n$

  • Sigma ($\Sigma$) Notation Anatomy: $$\sum_{k=p}^{q} (ak + b)$$
    • Number of terms $N$: $N = q - p + 1$ (never just $q - p$!).
    • First term $U_1$: substitute lower limit $k = p$ into the expression: $U_1 = ap + b$.
    • Last term $U_N$: substitute upper limit $k = q$: $U_N = aq + b$.
    • Common difference: the coefficient of the index variable, which is $a$.
    • Sum: $S_N = \frac{N}{2}(U_1 + U_N)$.
  • Solving for the Number of Terms $n$ Given $S_n$: When the total sum $S_n$ is known and you need to find $n$, substitute into $S_n = \frac{n}{2}[2u_1 + (n - 1)d]$ to produce a quadratic equation in $n$: $$\frac{d}{2}n^2 + \left(u_1 - \frac{d}{2}\right)n - S_n = 0$$ Solve by factoring or the quadratic formula, and discard any negative or non-integer roots since $n \in \mathbb{Z}^+$.

Example 4

Finding $n$ by Forming and Solving a Quadratic Equation

An arithmetic sequence has first term $u_1 = 5$ and common difference $d = 3$. The sum of the first $n$ terms is $440$. Find the value of $n$.

  1. Set up the equation using $S_n = \frac{n}{2}[2u_1 + (n - 1)d]$: $$440 = \frac{n}{2}\left[ 2(5) + (n - 1)(3) \right]$$
  2. Multiply both sides by $2$ and expand: $$880 = n\left[ 10 + 3n - 3 \right] = n\left[ 3n + 7 \right]$$ $$3n^2 + 7n - 880 = 0$$
  3. Solve the quadratic equation using the quadratic formula: $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4(3)(-880)}}{2(3)}$$ $$\sqrt{49 + 10560} = \sqrt{10609} = 103$$ $$n = \frac{-7 \pm 103}{6}$$
  4. Evaluate roots and select the valid integer $n > 0$: $$n = \frac{96}{6} = 16 \quad \text{or} \quad n = \frac{-110}{6} \quad (\text{rejected})$$

Answer: $n = 16$

1.2

Recovering Terms and Checking the Model

  • Recover a term from totals: $u_n=S_n-S_{n-1}$ for $n\ge2$, and $u_1=S_1$. Define $S_0=0$ to use the same rule at $n=1$.
  • Example: If $S_n=2n^2+3n$, then $u_n=(2n^2+3n)-[2(n-1)^2+3(n-1)]=4n+1$. Hence $u_1=5$ and $d=4$.
  • Constant sequence: When $d=0$, every term equals $u_1$ and $S_n=nu_1$. The equation for $n$ is linear rather than quadratic.
  • Keep every valid root: Negative differences can give two positive integer term counts for the same total. For $u_1=10$, $d=-1$ and $S_n=45$, $(n-6)(n-15)=0$, so both $n=6$ and $n=15$ work.
  • Thresholds: For a positive, increasing total, solve the inequality and check the integers either side of the boundary. A strict requirement such as “exceeds” must not be replaced by “equals”.
  • Model limits: A constant increase is an assumption. If a model counts seats, products or people, terms and term positions must satisfy the real-world constraints.

Example 5

Consecutive Terms Parameter & Sum Evaluation

The terms $2k - 1$, $k + 4$, and $3k - 2$ are three consecutive terms of an arithmetic sequence.
(a) Find the value of $k$. [3 marks]
(b) Given that these are the 4th, 5th, and 6th terms of the sequence, find the first term $u_1$ and the sum of the first 20 terms $S_{20}$. [4 marks]

  1. Part (a): Apply the consecutive terms condition $2b = a + c$: $$2(k + 4) = (2k - 1) + (3k - 2) \quad [M1]$$ $$2k + 8 = 5k - 3 \quad [A1]$$ $$3k = 11 \implies k = \frac{11}{3} \quad [A1]$$
  2. Part (b): Find the common difference $d$ and first term $u_1$: $$u_4 = 2\left(\frac{11}{3}\right) - 1 = \frac{22}{3} - \frac{3}{3} = \frac{19}{3}$$ $$u_5 = \frac{11}{3} + 4 = \frac{11}{3} + \frac{12}{3} = \frac{23}{3}$$ $$d = u_5 - u_4 = \frac{23}{3} - \frac{19}{3} = \frac{4}{3} \quad [A1]$$ Use $u_4 = u_1 + 3d$: $$\frac{19}{3} = u_1 + 3\left(\frac{4}{3}\right) = u_1 + 4 \implies u_1 = \frac{19}{3} - \frac{12}{3} = \frac{7}{3} \quad [M1][A1]$$
  3. Calculate $S_{20}$ using $S_n = \frac{n}{2}[2u_1 + (n - 1)d]$: $$S_{20} = \frac{20}{2}\left[ 2\left(\frac{7}{3}\right) + 19\left(\frac{4}{3}\right) \right] = 10\left[ \frac{14 + 76}{3} \right] = 10\left[ \frac{90}{3} \right] = 10(30) = 300 \quad [A1]$$

Answers: (a) $k = \frac{11}{3}$; (b) $u_1 = \frac{7}{3}$, $S_{20} = 300$

Examiner Commentary: Candidates often compute $u_5 - u_4$ correctly but mistakenly treat $u_4$ as $u_1$. Always write down the general term index carefully: $u_4 = u_1 + 3d$.

Tier 1: Core Fluency (Questions 1–4)
Exercise 1 Paper 1 Core Term & Sum

An arithmetic sequence begins $8, 13, 18, 23, \dots$

(a) Find the common difference $d$ and write down the expression for $u_n$.

(b) Calculate the $20$th term $u_{20}$.

(c) Find the sum of the first $20$ terms $S_{20}$.

View Complete Solution
  1. Part (a): $u_1 = 8$, $d = 13 - 8 = 5$. $$u_n = 8 + (n - 1)(5) = 5n + 3$$
  2. Part (b): $$u_{20} = 5(20) + 3 = 103$$
  3. Part (c): $$S_{20} = \frac{20}{2}(u_1 + u_{20}) = 10(8 + 103) = 10(111) = 1110$$

Final Answer: (a) $d = 5, u_n = 5n + 3$; (b) $u_{20} = 103$; (c) $S_{20} = 1110$

Exercise 2 Paper 1 Decreasing AP Threshold

An arithmetic sequence has $u_1 = 92$ and $d = -4$. Find the first term of the sequence that is less than $0$, and state its position $n$.

View Complete Solution
  1. Formulate the $n$th term: $$u_n = 92 + (n - 1)(-4) = 92 - 4n + 4 = 96 - 4n$$
  2. Set up inequality $u_n < 0$: $$96 - 4n < 0 \implies 4n > 96 \implies n > 24$$
  3. Find the smallest integer $n$: Since $n > 24$, the first term below zero is $n = 25$.
    $$u_{25} = 96 - 4(25) = 96 - 100 = -4$$

Final Answer: The $25$th term ($u_{25} = -4$)

Exercise 3 Paper 1 Simultaneous System

In an arithmetic sequence, the $5$th term is $22$ and the $12$th term is $57$. Find the first term $u_1$ and the common difference $d$.

View Complete Solution
  1. Write simultaneous equations: $$u_1 + 4d = 22 \quad \text{--- (1)}$$ $$u_1 + 11d = 57 \quad \text{--- (2)}$$
  2. Subtract (1) from (2): $$7d = 35 \implies d = 5$$
  3. Find $u_1$: $$u_1 + 4(5) = 22 \implies u_1 + 20 = 22 \implies u_1 = 2$$

Final Answer: $u_1 = 2, d = 5$

Exercise 4 Paper 1 Sigma Evaluation

Evaluate the finite series written in sigma notation:

$$\sum_{k=1}^{30} (3k + 2)$$

View Complete Solution
  1. Identify parameters: Number of terms $n = 30 - 1 + 1 = 30$.
    First term ($k = 1$): $u_1 = 3(1) + 2 = 5$.
    Last term ($k = 30$): $u_{30} = 3(30) + 2 = 92$.
  2. Use series formula $S_n = \frac{n}{2}(u_1 + u_n)$: $$S_{30} = \frac{30}{2}(5 + 92) = 15(97) = 1455$$

Final Answer: $1455$

Tier 2: Exam Modeling (Questions 5–7)
Exercise 5 Checking a Proposed Total

The first term of an arithmetic sequence is $4$ and the common difference is $3$. Given that the sum of the first $n$ terms is $435$, find the value of $n$.

View Complete Solution
  1. Form the sum equation: $$435 = \frac{n}{2}[8 + 3(n-1)] \implies 3n^2 + 5n - 870 = 0.$$
  2. Check the permitted values of $n$: $$n = \frac{-5 \pm \sqrt{10465}}{6}.$$ Since $102^2 < 10465 < 103^2$, the square root is not an integer. Neither root is a positive integer, so neither can count terms.
  3. Confirm using neighbouring partial sums: $$S_{16}=\frac{16}{2}[8+15(3)]=424,\qquad S_{17}=\frac{17}{2}[8+16(3)]=476.$$ Every term is positive, so the partial sums increase strictly. The proposed total $435$ lies between consecutive partial sums and cannot occur.

Final Answer: No positive integer $n$ satisfies the stated total. The supplied data are inconsistent; rounding a root or changing the total does not solve the original question.

Exercise 6 Paper 1 Consecutive Terms Algebraic Parameter

The terms $2x + 1$, $x^2 - 1$, and $5x + 3$ are three consecutive terms of an arithmetic sequence, in that order. Find all possible values of $x$.

View Complete Solution
  1. Apply the arithmetic mean condition $2b = a + c$: $$2(x^2 - 1) = (2x + 1) + (5x + 3)$$
  2. Expand and collect like terms: $$2x^2 - 2 = 7x + 4$$ $$2x^2 - 7x - 6 = 0$$
  3. Solve the quadratic equation: $$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-6)}}{2(2)} = \frac{7 \pm \sqrt{49 + 48}}{4} = \frac{7 \pm \sqrt{97}}{4}$$

Final Answer: $x = \frac{7 + \sqrt{97}}{4} \quad \text{and} \quad x = \frac{7 - \sqrt{97}}{4}$

Exercise 7 Paper 1 Non-Unity Sigma Index

Evaluate the sum:

$$\sum_{k=6}^{25} (4k - 7)$$

View Complete Solution
  1. Calculate the number of terms $N$: $$N = \text{upper} - \text{lower} + 1 = 25 - 6 + 1 = 20\text{ terms}$$
  2. Find first term ($k = 6$) and last term ($k = 25$): $$u_1 = 4(6) - 7 = 24 - 7 = 17$$ $$u_{20} = 4(25) - 7 = 100 - 7 = 93$$
  3. Apply series formula $S_N = \frac{N}{2}(u_1 + u_N)$: $$S = \frac{20}{2}(17 + 93) = 10(110) = 1100$$

Final Answer: $1100$

Tier 3: Complex Applications & IB Exam Modeling (Questions 8–10)
Exercise 8 Paper 2 Contextual Seating Model

An open-air amphitheatre has $20$ seats in the first row. Each successive row has $3$ more seats than the row directly in front of it. The amphitheatre contains $30$ rows in total.

(a) Find the number of seats in the $30$th row.

(b) Calculate the total seating capacity of the amphitheatre.

(c) Each ticket in the first $10$ rows costs $\$45$, while each ticket in the remaining $20$ rows costs $\$30$. Assuming a sold-out performance, calculate the total ticket revenue.

View Complete Solution
  1. Part (a): $u_1 = 20, d = 3, n = 30$: $$u_{30} = 20 + 29(3) = 20 + 87 = 107\text{ seats}$$
  2. Part (b): Total capacity $S_{30}$: $$S_{30} = \frac{30}{2}(20 + 107) = 15(127) = 1905\text{ seats}$$
  3. Part (c): Split into two tier sums: First $10$ rows: $S_{10} = \frac{10}{2}[2(20) + 9(3)] = 5(40 + 27) = 5(67) = 335\text{ seats}$.
    Remaining $20$ rows: $1905 - 335 = 1570\text{ seats}$.
    $$\text{Total Revenue} = (335 \times \$45) + (1570 \times \$30) = \$15\,075 + \$47\,100 = \$62\,175$$

Final Answer: (a) $107$ seats; (b) $1905$ seats; (c) $\$62\,175$

Exercise 9 Paper 1 Consecutive Block Sums

In an arithmetic sequence, the sum of the first $8$ terms is $140$, and the sum of the next $8$ terms (from the $9$th to the $16$th term) is $396$. Find the first term $u_1$ and the common difference $d$.

View Complete Solution
  1. Use the first sum condition $S_8 = 140$: $$S_8 = \frac{8}{2}(2u_1 + 7d) = 4(2u_1 + 7d) = 140 \implies 2u_1 + 7d = 35 \quad \text{--- (1)}$$
  2. Calculate the sum of all first $16$ terms: $$S_{16} = S_8 + (\text{next } 8\text{ terms}) = 140 + 396 = 536$$
  3. Set up equation for $S_{16}$: $$S_{16} = \frac{16}{2}(2u_1 + 15d) = 8(2u_1 + 15d) = 536 \implies 2u_1 + 15d = 67 \quad \text{--- (2)}$$
  4. Subtract (1) from (2): $$(2u_1 + 15d) - (2u_1 + 7d) = 67 - 35 \implies 8d = 32 \implies d = 4$$
  5. Substitute $d = 4$ into (1): $$2u_1 + 7(4) = 35 \implies 2u_1 + 28 = 35 \implies 2u_1 = 7 \implies u_1 = 3.5 \quad \left(\text{or } \frac{7}{2}\right)$$

Final Answer: $u_1 = 3.5, d = 4$

Exercise 10 Paper 1 Logarithmic Arithmetic Progression

An arithmetic sequence has first term $u_1 = \ln a$ and common difference $d = \ln 2$, where $a > 0$.

(a) Write down an expression for $u_4$ in the form $\ln(ka)$, where $k \in \mathbb{Z}^+$.

(b) Show that the sum of the first $n$ terms can be expressed as $S_n = n\ln a + \frac{n(n - 1)}{2}\ln 2$.

(c) Given that $a = 8$ and $S_5 = p\ln 2$, find the integer value of $p$.

View Complete Solution
  1. Part (a): Find $u_4$: $$u_4 = u_1 + 3d = \ln a + 3\ln 2 = \ln a + \ln(2^3) = \ln a + \ln 8 = \ln(8a)$$ Hence $k = 8$.
  2. Part (b): Apply sum formula: $$S_n = \frac{n}{2}[2u_1 + (n - 1)d] = \frac{n}{2}[2\ln a + (n - 1)\ln 2] = n\ln a + \frac{n(n - 1)}{2}\ln 2 \quad \text{[Shown]}$$
  3. Part (c): Substitute $a = 8$ and $n = 5$: Since $a = 8 = 2^3$, $\ln a = \ln(2^3) = 3\ln 2$.
    $$S_5 = 5(3\ln 2) + \frac{5(4)}{2}\ln 2 = 15\ln 2 + 10\ln 2 = 25\ln 2$$ Therefore, $p = 25$.

Final Answer: (a) $u_4 = \ln(8a)$; (b) [Shown]; (c) $p = 25$

Exercise 11Independent Application

A workshop makes $48$ lanterns in week 3 and $72$ lanterns in week 7. It plans to increase production by the same number each week. It has material for $900$ lanterns in total. Determine the greatest number of complete weeks it can operate without exceeding that supply, the material remaining, and one limitation of the model. Choose your own method.

View Complete Solution
  1. Let $a$ be week 1 production and $d$ the weekly increase. Then $a+2d=48$ and $a+6d=72$, so $d=6$ and $a=36$.
  2. The total after $n$ weeks is $S_n=\frac{n}{2}[72+6(n-1)]=3n^2+33n$. Require $S_n\le900$, giving $n^2+11n-300=(n+25)(n-12)\le0$.
  3. Since $n$ is a positive integer, the maximum is $12$. Check $S_{12}=900$ and $S_{13}=936$.
  4. Material remaining: $900-900=0$ lanterns' worth. A constant weekly increase may fail because of staffing, equipment capacity or changing demand; material for every lantern must also remain constant.

Answer: $12$ complete weeks; no material remaining. Any relevant, explained limitation is acceptable.

Formula Booklet SL 1.2

General Term & Difference

  • Common Difference: $$d = u_{k+1} - u_k$$
  • The $n$th Term: $$u_n = u_1 + (n - 1)d$$
  • Consecutive Terms (Arithmetic Mean): $$2u_k = u_{k-1} + u_{k+1}$$
Formula Booklet SL 1.2

Sum of an Arithmetic Series

  • Form with Known Last Term: $$S_n = \frac{n}{2}(u_1 + u_n)$$
  • Form with Known Difference: $$S_n = \frac{n}{2}\left[2u_1 + (n - 1)d\right]$$
  • Sigma Number of Terms: $$N = \text{upper} - \text{lower} + 1$$
Must Memorize

Exam Method Protocols

  • Difference Between Arbitrary Terms: $$u_q - u_p = (q - p)d$$
  • Recovering $n$th Term from Sums: $$u_n = S_n - S_{n-1} \quad (n \ge 2)$$
  • Sum Between Bounds: $$\sum_{k=p}^{q} u_k = S_q - S_{p-1}$$

Critical IB Examination Pitfalls in Arithmetic Sequences

  • Using $n$ Instead of $(n - 1)$: Writing $u_n = u_1 + nd$ is the single most common student error in Paper 1. Always write $(n - 1)$.
  • The Off-by-One Sigma Error: Evaluating $\sum_{k=4}^{20}$ with $n = 20 - 4 = 16$ terms instead of $20 - 4 + 1 = 17$ terms gives an incorrect term count and sum. Show your working so the method can be assessed.
  • Confusing $u_n$ with $S_n$: If a question states "the total amount over 12 months is $\$5000$," this is $S_{12} = 5000$, NOT $u_{12} = 5000$.
  • Sign Errors with Negative Common Differences: When $d = -3$, $(n - 1)(-3) = -3n + 3$. Forgetting to distribute the negative to the $+3$ ruins the sequence formula.
  • Discarding Negative Roots in Quadratics: When solving for $n$ from $S_n$, always explicitly state that $n \in \mathbb{Z}^+$ to justify rejecting a negative root.

Section A: Diagnostic Check

4 core concept check questions. Immediate feedback and scoring.

Score: 0 / 4

Question 1

What is the $15$th term of an arithmetic sequence that starts $6, 11, 16, 21, \dots$?

Correct! $u_1 = 6, d = 5$. Thus $u_{15} = 6 + (15 - 1)(5) = 6 + 14(5) = 6 + 70 = 76$.

Question 2

In an arithmetic sequence, $u_3 = 14$ and $u_9 = 38$. What is the common difference $d$?

Correct! $u_9 - u_3 = (9 - 3)d \implies 38 - 14 = 6d \implies 24 = 6d \implies d = 4$.

Question 3

How many terms are there in the series $\sum_{k=7}^{32} (2k + 1)$?

Correct! The number of terms is $\text{upper} - \text{lower} + 1 = 32 - 7 + 1 = 26$.

Question 4

If the sum of the first $n$ terms is $S_n = 3n^2 + 2n$, what is the first term $u_1$ and the second term $u_2$?

Correct! $u_1 = S_1 = 3(1)^2 + 2(1) = 5$. $S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$. Thus $u_2 = S_2 - S_1 = 16 - 5 = 11$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

An arithmetic sequence has $u_1 = 5$ and $u_8 = 47$.

(a) Show that the common difference is $d = 6$. [2 marks]
(b) Find the sum of the first $30$ terms $S_{30}$. [2 marks]
(c) Given that the sum of the first $k$ terms is $616$, find the value of $k$. [2 marks]
View Markscheme
(a)
$u_8 = u_1 + 7d \implies 47 = 5 + 7d$ M1
$7d = 42 \implies \mathbf{d = 6}$ A1

(b)
$S_{30} = \frac{30}{2}[2(5) + 29(6)] = 15[10 + 174]$ M1
$= 15(184) = \mathbf{2760}$ A1

(c)
$616 = \frac{k}{2}[10 + (k-1)6] = 3k^2+2k$ M1
$3k^2+2k-616=(3k+44)(k-14)=0$.
Reject $k=-\frac{44}{3}$ because a term count must be a positive integer. Thus $\mathbf{k=14}$; substitution gives $S_{14}=588+28=616$. A1
Paper 2 (Calculator Allowed) [6 marks]

A runner starts a marathon training program. On Day 1, she runs $3.5\text{ km}$. Each subsequent day, she increases her daily running distance by $0.75\text{ km}$.

(a) Find the distance she runs on Day 28. [2 marks]
(b) Calculate the total cumulative distance she covers over the first $28$ days of training. [2 marks]
(c) Find the minimum number of days of training required for her cumulative distance to exceed $500\text{ km}$. [2 marks]
View Markscheme
(a)
$u_{28} = u_1 + 27d = 3.5 + 27(0.75)$ M1
$= 3.5 + 20.25 = \mathbf{23.75\text{ km}}$ A1

(b)
$S_{28} = \frac{28}{2}(u_1 + u_{28}) = 14(3.5 + 23.75)$ M1
$= 14(27.25) = \mathbf{381.5\text{ km}}$ A1

(c)
$S_n = \frac{n}{2}[2(3.5) + (n - 1)(0.75)] > 500$ M1
$n(7 + 0.75n - 0.75) > 1000 \implies 0.75n^2 + 6.25n - 1000 > 0$
The positive boundary is $n=\frac{-6.25+\sqrt{3039.0625}}{1.5}\approx 32.5851$. Check the integers: $S_{32}=484$ km and $S_{33}=511.5$ km. Thus the minimum is $\mathbf{33\text{ days}}$. A1

Unit Overview

All Lessons in Number & Algebra (Topic 1)

1.1 Standard Form Scientific notation, powers-of-ten operations, and the IB 3 s.f. rule. 1.2 Arithmetic Sequences & Series Common difference, general $n$th term, and partial sum formulas. 1.3, 1.8 Geometric Sequences & Infinite Sums Common ratio, finite sums, and convergence to infinity when $|r| < 1$. 1.4 Financial Applications Compound interest, annual depreciation, and TVM solver modeling. 1.5, 1.7 Exponents & Logarithms Laws of exponents, logarithm rules, base changes, and natural logs. 1.6, 1.9 Proof & Binomial Theorem Deductive proof, Pascal's triangle, and binomial expansion terms.