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  • SL 1.2 : Arithmetic Sequences and Series

    Content Guidance, clarification and syllabus links
    Arithmetic sequences and series.

    Use of the formulae for the nth term and the sum of the first n terms of the sequence.

    Use of sigma notation for sums of arithmetic sequences.     		Spreadsheets, GDCs and graphing software may be used to generate and display sequences in several ways.

    If technology is used in examinations, students will be expected to identify the first term and the common difference.

    Applications include simple interest over a number of years.

    Analysis, interpretation and prediction where a model is not perfectly arithmetic in real life. Students will need to approximate common differences.

    📌 Introduction

    Arithmetic sequences and series are fundamental patterns in mathematics where consecutive terms have a constant difference. These sequences model real-world phenomena such as simple interest calculations, linear growth patterns, and regular payment schedules. Understanding arithmetic sequences provides the foundation for analyzing linear relationships and making predictions based on constant rate patterns.

    The beauty of arithmetic sequences lies in their predictability – once you know the first term and the common difference, you can find any term in the sequence or calculate the sum of any number of consecutive terms. This makes them incredibly useful for modeling situations where there is steady, linear growth or decline over time.

    📌 Definition Table

    Term Definition
    Arithmetic Sequence A sequence where the difference between consecutive terms is constant.
    Example: 2, 5, 8, 11, 14, … (common difference = 3)
    Common Difference (d) The constant value added to each term to get the next term: \(d = u_{n+1} – u_n\)
    Can be positive, negative, or zero
    First Term (a or \(u_1\)) The initial term of the sequence, often denoted as \(u_1\) or \(a\)
    nth Term (\(u_n\)) General term of the sequence: \(u_n = a + (n-1)d\)
    Also written as: \(u_n = u_1 + (n-1)d\)
    Arithmetic Series The sum of terms in an arithmetic sequence
    Example: 2 + 5 + 8 + 11 + 14 = 40
    Sum of n terms (\(S_n\)) \(S_n = \frac{n}{2}[2a + (n-1)d]\) or \(S_n = \frac{n}{2}(u_1 + u_n)\)
    Sigma Notation (\(\sum\)) Compact notation for expressing the sum of a series
    \(\sum_{k=1}^{n} u_k = \sum_{k=1}^{n} [a + (k-1)d]\)

    📌 Properties & Key Formulas

    • General Form: \(u_1, u_1 + d, u_1 + 2d, u_1 + 3d, \ldots\)
    • nth Term Formula: \(u_n = a + (n-1)d\) where \(a\) is first term, \(d\) is common difference
    • Sum Formula (Method 1): \(S_n = \frac{n}{2}[2a + (n-1)d]\)
    • Sum Formula (Method 2): \(S_n = \frac{n}{2}(u_1 + u_n)\)
    • Alternative Sum Formula: \(S_n = \frac{n}{2}(\text{first term} + \text{last term})\)
    • Sigma Notation: \(\sum_{k=1}^{n} u_k = \sum_{k=1}^{n} [a + (k-1)d]\)
    • Key Property: In an arithmetic sequence, \(u_n = \frac{u_{n-1} + u_{n+1}}{2}\) (middle term is average of neighbors)

    Common Examples:

    • \(2, 5, 8, 11, 14, \ldots\) (first term: 2, common difference: 3)
    • \(10, 7, 4, 1, -2, \ldots\) (first term: 10, common difference: -3)
    • \(5, 5, 5, 5, 5, \ldots\) (first term: 5, common difference: 0)
    • \(-3, 1, 5, 9, 13, \ldots\) (first term: -3, common difference: 4)
    🧠 Examiner Tip: Always identify the first term and common difference clearly before applying formulas.

    Check your common difference by testing multiple consecutive pairs of terms. Remember that \(d = u_2 – u_1 = u_3 – u_2 = u_4 – u_3\), etc.

    📌 Common Mistakes & How to Avoid Them

    ⚠️ Common Mistake #1: Confusing \(n\) and \(u_n\)

    Wrong: “The 5th term is 5” when actually \(u_5 = 17\)
    Right: “\(n = 5\) (position), \(u_5 = 17\) (value)”

    How to avoid: Always clearly distinguish between the position of a term and its value.
    ⚠️ Common Mistake #2: Wrong formula substitution

    Wrong: Using \(u_n = a + nd\) instead of \(u_n = a + (n-1)d\)
    Example: For sequence 3, 7, 11, 15, … finding \(u_4\)
    Wrong: \(u_4 = 3 + 4 \times 4 = 19\) ❌
    Right: \(u_4 = 3 + (4-1) \times 4 = 3 + 12 = 15\) ✅

    How to avoid: Remember the formula has \((n-1)\) because we start counting from the first term.
    ⚠️ Common Mistake #3: Sign errors with negative common differences

    Wrong: For sequence 20, 15, 10, 5, … claiming \(d = 5\)
    Right: \(d = 15 – 20 = -5\)

    How to avoid: Always subtract in the correct order: \(d = u_{n+1} – u_n\)
    ⚠️ Common Mistake #4: Rounding too early in calculations

    Wrong: Rounding intermediate steps and getting incorrect final answers
    Right: Keep full precision until the final step

    How to avoid: Use calculator memory functions or keep extra decimal places during calculations.
    ⚠️ Common Mistake #5: Mixing up sum formulas

    Wrong: Using \(S_n = \frac{n}{2}[2a + nd]\)
    Right: \(S_n = \frac{n}{2}[2a + (n-1)d]\)

    How to avoid: Double-check your formula before substituting values.

    📌 Calculator Skills: Casio CG-50 & TI-84

    📱 Using Casio CG-50 for Arithmetic Sequences

    Method 1: Using TABLE function
    1. Press [MENU] → Select “Graph”
    2. Enter your sequence formula as Y1 = 3X + 2 (for \(u_n = 3n + 2\))
    3. Press [F6] (TABLE) to generate terms
    4. Use [SET] to adjust starting value and step size

    Method 2: Using List function
    1. Press [MENU] → Select “Run-Matrix”
    2. Press [OPTN] → [LIST] → [SEQ]
    3. Enter: seq(3X+2, X, 1, 10, 1) for first 10 terms
    4. Press [EXE] to calculate

    Finding specific terms:
    • In TABLE view, enter the desired n-value in X column
    • Calculator shows corresponding term value

    Calculating sums:
    1. Create list of terms using seq() function
    2. Press [OPTN] → [LIST] → [SUM]
    3. Apply to your list: sum(Ans)
    📱 Using TI-84 for Arithmetic Sequences

    Setting up sequence mode:
    1. Press [MODE], scroll to “SEQ” and press [ENTER]
    2. Press [Y=] to access sequence editor

    Entering sequences:
    1. In Y= menu, enter: u(n) = 3n + 2
    2. Set nMin = 1 (starting value)
    3. Press [2nd] → [WINDOW] to set sequence parameters

    Generating terms from home screen:
    1. Press [2nd] → [STAT] → [OPS] → [5:seq(]
    2. Enter: seq(3X+2, X, 1, 10, 1)
    3. Press [ENTER] for first 10 terms

    Finding sums:
    1. Press [2nd] → [STAT] → [MATH] → [5:sum(]
    2. Enter: sum(seq(3X+2, X, 1, 10, 1))
    3. Press [ENTER] to calculate sum

    Individual terms:
    • After setting up sequence in Y=, use u(20) to find 20th term
    📱 Calculator Tips & Tricks

    Verification method:
    • Always check first few terms manually to verify your formula is correct
    • Use calculator to find pattern, then verify with hand calculations

    Avoiding errors:
    • Use parentheses: (n-1) not n-1 in formulas
    • For negative common difference, use (-) not subtract key
    • Store intermediate results in calculator memory

    Graphing sequences:
    • Both calculators can plot sequences graphically
    • Use TIME plot for discrete points
    • Helpful for visualizing convergence or growth patterns

    📌 Mind Map

    📌 Applications in Science and IB Math

    • Finance & Economics: Simple interest calculations, linear depreciation, regular savings plans, salary increments
    • Physics & Engineering: Uniform acceleration problems, linear motion, step functions in electrical circuits
    • Biology & Medicine: Drug dosage schedules, population growth in controlled environments, linear enzyme kinetics
    • Chemistry: Concentration changes in linear dilution processes, temperature changes at constant rates
    • Computer Science: Algorithm analysis, array indexing, memory allocation patterns
    • Statistics: Linear regression models, trend analysis in data sets
    • Architecture & Construction: Staircase design, regular spacing in structural elements
    • Environmental Science: Linear pollution accumulation, regular monitoring intervals
    ➗ IA Tips & Guidance: Use arithmetic sequences to model real-world linear growth patterns in your IA.

    Excellent IA Topics:
    • Salary progression analysis in different career paths
    • Simple vs compound interest comparison over time
    • Linear depreciation of vehicle values
    • Population growth in controlled laboratory conditions
    • Temperature change patterns in cooling/heating processes

    IA Structure Tips:
    • Collect real data and fit arithmetic models
    • Compare theoretical vs actual results
    • Discuss limitations of linear models
    • Use technology to generate and analyze large datasets

    📌 Worked Examples (IB Style)

    Q1. Find the 15th term of the arithmetic sequence: 3, 7, 11, 15, …

    Solution:

    Step 1: Identify the first term and common difference
    First term: \(a = u_1 = 3\)
    Common difference: \(d = 7 – 3 = 4\)

    Step 2: Apply the nth term formula
    \(u_n = a + (n-1)d\)

    Step 3: Substitute values
    \(u_{15} = 3 + (15-1) \times 4 = 3 + 14 \times 4 = 3 + 56 = 59\)

    ✅ Final answer: \(u_{15} = 59\)

    Q2. Find the sum of the first 20 terms of the sequence: 5, 9, 13, 17, …

    Solution:

    Step 1: Identify values
    \(a = 5\), \(d = 9 – 5 = 4\), \(n = 20\)

    Step 2: Choose appropriate sum formula
    Using \(S_n = \frac{n}{2}[2a + (n-1)d]\)

    Step 3: Substitute and calculate
    \(S_{20} = \frac{20}{2}[2(5) + (20-1)(4)]\)
    \(= 10[10 + 19 \times 4]\)
    \(= 10[10 + 76] = 10 \times 86 = 860\)

    ✅ Final answer: \(S_{20} = 860\)

    Alternative Method:
    First find \(u_{20} = 5 + 19 \times 4 = 81\)
    Then use \(S_{20} = \frac{20}{2}(5 + 81) = 10 \times 86 = 860\)

    Q3. Express the sum \(3 + 7 + 11 + 15 + \ldots + 47\) using sigma notation and find its value.

    Solution:

    Step 1: Find the general term
    \(a = 3\), \(d = 7 – 3 = 4\)
    \(u_n = 3 + (n-1) \times 4 = 3 + 4n – 4 = 4n – 1\)

    Step 2: Find which term equals 47
    \(4n – 1 = 47\)
    \(4n = 48\)
    \(n = 12\)

    Step 3: Write in sigma notation
    \(\sum_{k=1}^{12} (4k – 1)\)

    Step 4: Calculate the sum
    \(S_{12} = \frac{12}{2}(3 + 47) = 6 \times 50 = 300\)

    ✅ Final answer: \(\sum_{k=1}^{12} (4k – 1) = 300\)

    Q4. A salary starts at $40,000 and increases by $2,500 each year. Find the total earnings over 10 years.

    Solution:

    Step 1: Identify the arithmetic sequence
    This represents an arithmetic sequence where:
    \(a = 40000\) (first year salary)
    \(d = 2500\) (annual increase)
    \(n = 10\) (number of years)

    Step 2: Apply sum formula
    \(S_{10} = \frac{10}{2}[2(40000) + (10-1)(2500)]\)

    Step 3: Calculate step by step
    \(S_{10} = 5[80000 + 9 \times 2500]\)
    \(= 5[80000 + 22500]\)
    \(= 5 \times 102500 = 512500\)

    ✅ Final answer: Total earnings = $512,500

    Real-world interpretation:
    • Year 1: $40,000
    • Year 2: $42,500
    • Year 3: $45,000
    • …
    • Year 10: $40,000 + 9 × $2,500 = $62,500

    Q5. Find the value of x if the terms 2x, x+3, and x-1 form an arithmetic sequence.

    Solution:

    Step 1: Use the property of arithmetic sequences
    For three consecutive terms in arithmetic sequence: \(u_2 – u_1 = u_3 – u_2\)
    Or equivalently: \(u_2 = \frac{u_1 + u_3}{2}\)

    Step 2: Set up the equation
    Common difference must be constant:
    \(d_1 = (x+3) – 2x = 3-x\)
    \(d_2 = (x-1) – (x+3) = -4\)

    Step 3: Solve for x
    Setting \(d_1 = d_2\):
    \(3-x = -4\)
    \(3+4 = x\)
    \(x = 7\)

    Step 4: Verify the answer
    When \(x = 7\): The sequence becomes 14, 10, 6
    Check: \(10 – 14 = -4\) and \(6 – 10 = -4\) ✓

    ✅ Final answer: \(x = 7\)

    📝 Paper Tip: Always verify your answers by checking if the calculated terms actually form an arithmetic sequence.

    Key verification steps:
    • Check that consecutive differences are equal
    • Substitute back into original conditions
    • Show clear working for finding first term and common difference
    • Use calculator to verify large calculations

    📌 Multiple Choice Questions (with Detailed Solutions)

    Q1. What is the 12th term of the arithmetic sequence 4, 9, 14, 19, …?

    A) 59     B) 54     C) 64     D) 49

    📖 Show Answer

    Step-by-step solution:

    1. Identify: \(a = 4\), \(d = 9 – 4 = 5\)

    2. Apply formula: \(u_n = a + (n-1)d\)

    3. Substitute: \(u_{12} = 4 + (12-1) \times 5 = 4 + 11 \times 5 = 4 + 55 = 59\)

    ✅ Answer: A) 59

    Q2. The sum of the first 8 terms of an arithmetic sequence is 108. If the first term is 6, what is the common difference?

    A) 2.5     B) 3     C) 3.5     D) 4

    📖 Show Answer

    Step-by-step solution:

    1. Given: \(S_8 = 108\), \(a = 6\), find \(d\)

    2. Use formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)

    3. Substitute: \(108 = \frac{8}{2}[2(6) + (8-1)d] = 4[12 + 7d]\)

    4. Solve: \(108 = 48 + 28d\)

    5. \(28d = 60\), so \(d = \frac{60}{28} = \frac{15}{7} ≈ 2.14\)

    6. Closest option is 2.5

    ✅ Answer: A) 2.5

    Q3. Which expression represents \(\sum_{k=1}^{10} (3k + 2)\)?

    A) 5 + 8 + 11 + … + 32     B) 3 + 6 + 9 + … + 30
    C) 2 + 5 + 8 + … + 29     D) 5 + 8 + 11 + … + 35

    📖 Show Answer

    Step-by-step solution:

    1. Find first term: when \(k=1\), \(3(1)+2=5\)

    2. Find second term: when \(k=2\), \(3(2)+2=8\)

    3. Find last term: when \(k=10\), \(3(10)+2=32\)

    4. The sequence is: 5, 8, 11, 14, 17, 20, 23, 26, 29, 32

    ✅ Answer: A) 5 + 8 + 11 + … + 32

    📌 Short Answer Questions (with Detailed Solutions)

    Q1. An arithmetic sequence has first term -3 and common difference 4. Find the first positive term of the sequence.

    📖 Show Answer

    Complete solution:

    Step 1: Write the general term

    \(u_n = -3 + (n-1) \times 4 = -3 + 4n – 4 = 4n – 7\)

    Step 2: Set up inequality for positive terms

    For first positive term: \(4n – 7 > 0\)

    \(4n > 7\)

    \(n > 1.75\)

    Step 3: Find smallest integer value

    Since \(n\) must be a positive integer, \(n = 2\)

    Step 4: Calculate the term

    \(u_2 = 4(2) – 7 = 8 – 7 = 1\)

    Verification: The sequence is: -3, 1, 5, 9, …

    ✅ Answer: The first positive term is 1

    Q2. The 5th term of an arithmetic sequence is 17 and the 12th term is 38. Find the first term and common difference.

    📖 Show Answer

    Complete solution:

    Step 1: Set up equations using \(u_n = a + (n-1)d\)

    \(u_5 = a + 4d = 17\) … (equation 1)

    \(u_{12} = a + 11d = 38\) … (equation 2)

    Step 2: Solve for common difference

    Subtract equation 1 from equation 2:

    \((a + 11d) – (a + 4d) = 38 – 17\)

    \(7d = 21\)

    \(d = 3\)

    Step 3: Find first term

    Substitute \(d = 3\) into equation 1:

    \(a + 4(3) = 17\)

    \(a + 12 = 17\)

    \(a = 5\)

    Verification: \(u_5 = 5 + 4(3) = 17\) ✓, \(u_{12} = 5 + 11(3) = 38\) ✓

    ✅ Answer: First term = 5, Common difference = 3

    📌 Extended Response Questions (with Full Solutions)

    Q1. A company’s annual profit forms an arithmetic sequence. In the first year, the profit was $50,000. In the fourth year, it was $65,000.

    (a) Find the common difference and write the general term.

    (b) Calculate the total profit for the first 10 years.

    (c) In which year will the profit first exceed $100,000?

    📖 Show Answer

    Complete solution:

    (a) Finding common difference and general term

    Given: \(u_1 = 50000\), \(u_4 = 65000\)

    Using \(u_4 = u_1 + 3d\):

    \(65000 = 50000 + 3d\)

    \(3d = 15000\)

    \(d = 5000\)

    General term: \(u_n = 50000 + (n-1) \times 5000 = 45000 + 5000n\)

    (b) Total profit for first 10 years

    Using sum formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)

    \(S_{10} = \frac{10}{2}[2(50000) + (10-1)(5000)]\)

    \(= 5[100000 + 9 \times 5000]\)

    \(= 5[100000 + 45000]\)

    \(= 5 \times 145000 = 725000\)

    (c) Year when profit first exceeds $100,000

    Set up inequality: \(45000 + 5000n > 100000\)

    \(5000n > 55000\)

    \(n > 11\)

    Therefore, in year 12: \(u_{12} = 45000 + 5000(12) = 105000\)

    Check year 11: \(u_{11} = 45000 + 5000(11) = 100000\) (exactly $100,000)

    Summary of yearly profits:
    Year 1: $50,000 | Year 2: $55,000 | Year 3: $60,000 | Year 4: $65,000
    Year 10: $95,000 | Year 11: $100,000 | Year 12: $105,000

    ✅ Final Answers:
    (a) Common difference = $5,000; General term: \(u_n = 45000 + 5000n\)
    (b) Total profit for 10 years = $725,000
    (c) Year 12 (profit = $105,000)

  • SL 1.1 : Understanding Scientific Notation



    Content Guidance, clarification and syllabus links
    Operations with numbers in the form \( a \times 10^k \) where \( 1 \leq a < 10 \) and \( k \) is an integer. Calculator or computer notation is not acceptable.
    For example, 5.2E30 is not acceptable and should be written as \( 5.2 \times 10^{30} \).

    📌 Introduction

    Standard form (\(a \times 10^k\)) is essential for efficiently representing very large or small numbers in mathematics and science.
    It allows for clear interpretation, maintains precision, and simplifies calculations. Commonly used in Astronomy, Physics, Chemistry, Biology and global finance, this notation avoids lengthy and error-prone digit strings.

    📌 Properties & Examples

    • Numbers must be written as \(a \times 10^k\) with \(1 \leq a < 10\), \(k \in \mathbb{Z}\).
    • \(5 \times 10^{3.2}\): Incorrect, \(k\) must be integer.
    • \(0.89 \times 10^{-4}\): Incorrect, \(a\) must be ≥ 1. Correct: \(8.9 \times 10^{-5}\).
    • \(2.3 \times 10^8\): Correct, equals 230,000,000.
    • Calculator notation such as 5.2E30 is not acceptable; use \(5.2 \times 10^{30}\).
    🧠 Examiner Tip: Always write numbers with \(a\) between 1 and 10, \(k\) an integer.
    Avoid calculator notation like E30.

    📌 Mind Map

    Mind map

    📌 Applications in Science and IB Math

    • Astronomical distances (e.g. \(1.5 \times 10^{11}\) m: Earth–Sun).
    • Microscopic measurements (e.g. \(3 \times 10^{-10}\) m: atom diameter).
    • Chemistry (Avogadro’s number: \(6.02 \times 10^{23}\)).
    • Physics (order of magnitude calculations, uncertainty).
    • Biology (cell sizes and concentrations).
    • Finance (global GDPs, populations).
    💡 IA Tips & Guidance: Use standard form for all calculated results requiring large/small numbers (widths, frequencies, concentrations).

    Explicitly state your rounding, and always convert to correct form for final answers.

    📌 Worked Examples (IB Style)

    Q1. Write 0.0000475 in standard form.

    Solution:

    \(0.0000475 = 4.75 \times 10^{-5}\)

    Here, \(a = 4.75\) is between 1 and 10, \(k = -5\) is an integer.

    Final answer: \(4.75 \times 10^{-5}\)

    Q2. Express 3,600,000,000 in standard form.

    Solution:

    \(3,600,000,000 = 3.6 \times 10^{9}\)

    Here, \(a = 3.6\) is between 1 and 10, \(k = 9\) is an integer.

    Final answer: \(3.6 \times 10^{9}\)

    Q3. Simplify \((2.5 \times 10^{4}) \times (3 \times 10^{6})\) and write the answer in standard form.

    Solution:

    \((2.5 \times 10^{4}) \times (3 \times 10^{6}) = (2.5 \times 3) \times (10^{4} \times 10^{6}) = 7.5 \times 10^{10}\)

    Here, \(a = 7.5\) is between 1 and 10, \(k = 10\) is an integer.

    Final answer: \(7.5 \times 10^{10}\)

    📝 Paper 2: IB questions use standard form throughout for large or small numbers in context.

    Always check the question requirements and final marking scheme for correct form—partial answers may lose marks if not properly rounded or written.

  • D4.3 CLIMATE CHANGE

  • D4.2 STABILITY AND CHANGE

  • D4.1 NATURAL SELECTION

  • D3.3 HOMEOSTASIS

  • D3.2 INHERITANCE

  • D3.1 REPRODUCTION

  • D2.3 WATER POTENTIAL

  • D2.2 GENE EXPRESSION [HL ONLY]