Unit 1

Patterns and Sequences

About this unit

Identify rules in number sequences, repeating patterns, and arithmetic progressions. Learn the cycle-remainder shortcut for repeating sequences and the sum formulas for arithmetic series.

What types of questions will you face?

  • 1Find the element at a large position in a repeating sequence using the cycle-remainder shortcut (divide position by cycle length; the remainder locates the answer in the cycle)
  • 2Calculate the nth term of an arithmetic sequence given the first term and common difference (nth term = first + (n−1) × difference)
  • 3Find the sum of an arithmetic series — such as all odd numbers from 1 to 99, or all multiples of 3 up to a limit — using the sum shortcut
  • 4Identify the rule in a non-linear or second-difference sequence and extend it step by step to find a later term
  • 5Work out the total number of items in a repeating multi-colour or multi-type pattern when you are told how many of one element there are (e.g. bead bracelets, knitting rows, page layouts)

Skills you will build

  • Applying the cycle-remainder method to pinpoint any term in a repeating pattern without listing every element
  • Writing and using the nth-term formula (a + (n−1)d) for arithmetic sequences quickly under time pressure
  • Calculating arithmetic series sums using the formula n ÷ 2 × (first + last) to avoid slow term-by-term addition
  • Spotting second-order differences (the gaps between gaps increase by a fixed amount) and extending sequences accurately
  • Translating a word-based pattern description into a structured cycle length and solving total-count problems from it

By the end of this unit, you will be able to

  • Find any term of a repeating sequence in seconds using division and remainder, without listing every element
  • Calculate any specific term or partial sum of an arithmetic sequence accurately and quickly
  • Identify whether a sequence is linear (arithmetic), exponential, or requires second-difference analysis — and apply the right method
  • Solve total-count problems for multi-element repeating patterns by working backwards from the given count of one element

Difficulty profile

Questions in this unit range from Very Easy to Medium. Repeating patterns and simple arithmetic sequences are Very Easy to Easy and reward knowing the shortcut. Second-difference sequences and triangle-pattern questions are Medium and appear in the mid-section of the OC MR paper.

Exam tip: Patterns and Sequences

For any repeating sequence, divide the position number by the cycle length and use only the remainder. If the remainder is 0, the answer is the last element of the cycle — not the first. For series sums, count the terms first using (last − first) ÷ step + 1, then apply n ÷ 2 × (first + last). Practise both reflexes until they are instant.

Sample Questions

Lesson 1 of 6Patterns and SequencesIntroductory

Let's begin with one of the most formulaic — and most reliably scorable — question types in OC Mathematical Reasoning: locating a specific term inside a repeating number or object sequence.

Repeating sequence questions appear in almost every OC MR paper, often placed in the opening third of the test. Students who master the one-step shortcut walk away with a quick, confident mark; those who count manually waste precious time and risk arithmetic errors.

The examiner wants to see whether you can recognise that a finite pattern simply wraps around and restarts, and that you can translate a large position number into a small, usable remainder — rather than physically listing every term up to that position.

A short sequence of 3–6 elements is shown repeating clearly (e.g. 13, 14, 15, 16, 13, 14, 15, 16, …). You are then asked to name the element at a large index — somewhere in the twenties, fifties, or hundreds — where counting every term would take far too long.

Best approach: Count the cycle length first. Divide the target position by the cycle length and keep only the remainder. The remainder directly gives you the position within the cycle. One edge case to remember: if the remainder is 0, the answer is the last element of the cycle — not the first.

Question

In the number pattern, 13, 14, 15, 16, 13, 14, 15, 16, 13, 14, 15, 16, …, what is the 34th number?

  1. A14
  2. B13
  3. C15
  4. D16
  5. E12

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Lesson 2 of 6Patterns and SequencesIntroductory

Ollie makes a necklace by repeating one step: 2 red beads then 3 blue beads. The finished necklace has 12 blue beads. How many red beads? The trick is that red and blue are added in a fixed ratio each step — so find how many steps ran, then multiply red beads per step by the number of steps.

Two-colour (or two-item) repeating pattern questions appear in the very easy to easy band of OC MR. They test ratio reasoning within a repeating cycle. The most common mistake is option A (6) — students divide 12 by 2 (the red count) instead of 12 by 3 (the blue count per step).

The examiner is testing whether students can identify the repeating unit (2 red, 3 blue), use the given total of one colour to find the number of repetitions, then apply that repetition count to find the total of the other colour. Option E (24 = 12 × 2) catches students who multiply directly without first finding the step count.

A pattern step adds fixed amounts of two different items (e.g. 2 red + 3 blue per cycle). The total of one item is given. Students must find the number of cycles and multiply by the per-cycle amount of the other item. All distractors arise from applying the wrong operation (dividing by wrong number, multiplying directly, subtracting).

Best approach: Step 1: Identify the repeating unit (2 red, 3 blue per step). Step 2: Use the known total to find the number of steps — total blue ÷ blue per step = 12 ÷ 3 = 4 steps. Step 3: Multiply red per step × number of steps = 2 × 4 = 8. Step 4: Verify both totals are consistent.

Question

Ollie makes a necklace by putting beads on a string.

He makes the necklace by repeating this step:

  • Add 2 red beads and then 3 blue beads onto the string.

The finished necklace has 12 blue beads.

How many red beads does it have?

  1. A6
  2. B8
  3. C11
  4. D18
  5. E24

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Lesson 3 of 6Patterns and SequencesEasy

The sequence 3, 6, 12, 24 doubles each time. The question asks for the difference between the 4th and 6th terms — so you need to generate two more terms first, then subtract. The most common trap is confusing the 6th term (96) for the answer, or finding the difference between the wrong pair of terms.

Doubling (geometric) sequences are among the most common sequence types in OC MR. Questions typically ask for a specific term, the sum of terms, or the difference between two terms. Generating extra terms by repeatedly doubling is a reliable and fast strategy.

The examiner tests whether students can extend a geometric sequence and then apply a secondary operation (subtraction). Option E (96) traps students who find the 6th term but forget to subtract the 4th. Option C (48) traps students who report the 5th term instead.

Four terms of a doubling sequence are given. Students must extend the sequence to the 5th and 6th terms, identify the 4th term (already given), then calculate the difference. The question is intentionally worded as '4th term and 6th term' to ensure students don't just subtract consecutive terms.

Best approach: Step 1: Spot the rule — each term is ×2. Step 2: Extend: 4th = 24, 5th = 48, 6th = 96. Step 3: Difference = 6th − 4th = 96 − 24 = 72.

Question

Here are the first four terms of a number sequence:

3, 6, 12, 24, …

What is the difference between the 4th term and the 6th term in this sequence?

  1. A12
  2. B24
  3. C48
  4. D72
  5. E96

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Lesson 4 of 6Patterns and SequencesIntermediate

Now let's step up to a more layered sequence type — one where the rule isn't immediately obvious. To crack it, you need to look one level deeper: at the gaps between consecutive terms rather than at the terms themselves.

Second-difference sequences appear regularly in the mid-to-upper difficulty range of OC MR tests. They are a reliable differentiator between students who stop at the surface pattern and those who know to investigate the gaps.

The examiner is checking whether you can identify a hidden arithmetic pattern in the differences between consecutive terms — specifically, whether those differences form their own arithmetic sequence — and then use that structure to extend the original sequence accurately.

You're given the first four or five terms of a sequence, sometimes with a hint about the gaps (e.g. 'differences are 3, 5, 7, 9, …'). In harder variants the hint is absent and you must discover it yourself. You're then asked for a term further along — typically the 6th or 7th.

Best approach: Write the sequence, then write the gap between each pair of consecutive terms directly below it. If those gaps form an arithmetic sequence (they increase or decrease by a fixed amount each step), simply extend the gaps one step at a time to find the next terms. Extending step by step is almost always faster and safer than attempting an nth-term formula.

Question

The first five terms of a sequence are:

2, 5, 10, 17, 26, …

The differences between terms are: 3, 5, 7, 9, … (odd numbers increasing by 2).

What is the 7th term in the sequence?

  1. A35
  2. B38
  3. C41
  4. D46
  5. E50

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Lesson 5 of 6Patterns and SequencesIntermediate

A sequence rule says each number is 4 less than double the previous. Jasper needs the number directly BEFORE 36, then adds its digits. The two-step trap: you must reverse the rule (work backwards from 36), not apply it forward. Option E (14) is what you get if you apply the rule forward to 36 — a classic direction error.

Rule-based sequence questions that require working backwards from a known term appear in the medium difficulty band of OC MR. They reward students who can algebraically reverse a rule (solve for P given next = 2P − 4) rather than generating terms forward from an unknown start.

The examiner is testing two skills: (1) reversing the sequence rule algebraically to find the predecessor of 36, and (2) adding the digits of the result. Option E (14) specifically traps students who apply the rule in the forward direction (2 × 36 − 4 = 68, then 6 + 8 = 14), confusing 'before' with 'after'.

A rule-based sequence is described (e.g. 'each number is 4 less than double the previous'). A specific term is given (36). Students must find the immediately preceding term by reversing the rule, then perform a secondary operation on it (add its digits, square it, etc.). One distractor comes from applying the rule forward instead of backward.

Best approach: Step 1: Write the rule algebraically — next = 2 × prev − 4. Step 2: Substitute the known term as 'next' and solve for 'prev': 2P − 4 = 36 → P = 20. Step 3: Verify — 2 × 20 − 4 = 36 ✓. Step 4: Apply the secondary operation — digits of 20 are 2 and 0, sum = 2.

Question

In a sequence, each number is 4 less than double the previous number.

Jasper finds the number directly before 36 in the sequence.

His answer has two digits. He adds these two digits to get a new number.

What is the new number?

  1. A2
  2. B4
  3. C5
  4. D7
  5. E14

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Lesson 6 of 6Patterns and SequencesIntermediate

Start at 53, apply 'odd→+1, even→÷2' five times: 53→54→27→28→14→7. The sequence alternates between adding 1 (to make odd numbers even) and halving (to reduce even numbers). Students who apply the rule only 4 times land on 14 (option C-ish) — counting steps carefully is critical.

Conditional rule sequences (if odd/even, do X/Y) applied a fixed number of times appear at the medium level of OC MR. Students must track the current value and parity at each step. The number of steps is small enough to trace manually but large enough to trip up careless students.

The examiner tests systematic application of a two-branch rule over exactly 5 steps. Common errors: applying the rule only 4 times (stopping at 14), or applying the wrong branch at one step.

A starting number and a conditional rule are given. Students apply the rule a specified number of times, tracking the value at each step.

Best approach: Trace each step: 53(odd)+1=54 → 54(even)÷2=27 → 27(odd)+1=28 → 28(even)÷2=14 → 14(even)÷2=7. Answer: 7.

Question

Shawn starts with the number 53 and follows this instruction five times:

  • If the number you have is odd, add 1, or if it is even, divide by 2.

After following the instruction five times, what number does Shawn have?

  1. A4
  2. B7
  3. C10
  4. D15
  5. E20

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