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 Selective 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 7Patterns and SequencesIntroductory

2, 6, ?, 54, 162 — check whether you add or multiply from one term to the next. 6 ÷ 2 = 3 and 162 ÷ 54 = 3, so every term is ×3. The missing term is just 6 × 3.

Geometric (multiply-by-constant) sequences appear in the very easy band of Selective MR. They’re quick marks for students who check ratios rather than differences, and a trap for those who assume every pattern is arithmetic (adding).

The examiner checks whether students recognise a ×3 geometric pattern rather than an arithmetic one (+4 between 2 and 6 is a red herring), and simply multiply the previous term by 3.

A number sequence with one term replaced by a symbol. The pattern is geometric (multiply by a constant). Students find the missing term.

Best approach: Divide consecutive known terms to find the ratio: 6÷2=3, 162÷54=3. Confirm ratio=3. Missing term = 6×3=18.

Question

What is the value of the missing number ▲ in this number pattern?

2, 6, ▲, 54, 162, …

What is the value of ▲?

  1. A10
  2. B12
  3. C18
  4. D30
  5. E50

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Lesson 2 of 7Patterns and SequencesEasy

Repeating and growing patterns on Selective MR often ask for the 30th or 100th item — never count one-by-one when you can find which block it lands in.

Bead- and object-pattern questions appear regularly in the easy band of Selective MR — students who spot how group sizes grow save time and avoid miscounts.

The examiner checks whether you can partition the pattern into labelled groups, track cumulative counts, and identify which group contains the target position.

A visual pattern alternates colours with group sizes 1, 2, 3, 4, … (or a fixed cycle repeats). You find the colour or symbol at a large index such as the 30th bead.

Best approach: List group number, colour, size, and running total until the total reaches or passes the target. The answer is the colour of the group that contains that position.

Question

What colour is the 30th bead in the pattern below?

● ○ ○ ● ● ● ○ ○ ○ ○ ● ● ● ● ● ○ ○ ○ ○ ○ ○ …

(● = grey bead, ○ = white bead)

  1. AWhite
  2. BGrey
  3. CCannot be determined
  4. DGrey, because 30 is even
  5. EWhite, because 30 ÷ 6 = 5

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Lesson 3 of 7Patterns and SequencesIntermediate

Rule-based sequences make you apply the same if-then rule step after step — write each new number before moving on so you do not skip a transformation.

Even/odd divide-or-multiply chains appear in the medium band of Selective MR — one reliable mark when you track parity at every step.

The examiner tests whether you can apply conditional rules consistently (even → halve, odd → ×3+1) across several iterations without losing your place.

You start from a given whole number and apply a two-branch rule a fixed number of times (e.g. four steps). You report the final value.

Best approach: After each step, note the number and whether it is even or odd, then apply exactly one branch. A tiny flow line (14 → 7 → 22 → …) prevents errors.

Question

Hayden starts with 14 and applies this rule four times:

  • If the number is even, divide by 2.
  • If the number is odd, multiply by 3 and add 1.

What number does he end up with?

  1. A34
  2. B22
  3. C11
  4. D17
  5. E40

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Lesson 4 of 7Patterns and SequencesIntermediate

Each child balances for half as long as the one before. Write out all five times first, then add carefully — convert to seconds to avoid decimal traps.

Halving (geometric) sequences appear regularly in Selective MR, often disguised as real-world scenarios (time, distance, money). The key skill is computing the sequence terms accurately and summing them without mistaking decimals for minutes and seconds.

The examiner tests whether students can apply a halving rule across 5 terms, convert 0.75 min to 45 seconds correctly (not 75 seconds), and add mixed minutes-and-seconds quantities. The most common error is treating 0.75 minutes as 75 seconds.

A sequence is defined by a starting value and a halving rule. Students must compute all terms and find their sum, expressing the answer in minutes and seconds.

Best approach: List all 5 terms: 12, 6, 3, 1.5, 0.75 minutes. Convert everything to seconds: 720 + 360 + 180 + 90 + 45 = 1395 seconds. Divide: 1395 ÷ 60 = 23 remainder 15 → 23 minutes 15 seconds.

Question

Five children take it in turn to do a balancing challenge, with each child completing it once.

The first child balances for 12 minutes.

After that, each child balances for half as long as the previous child.

What is the total time the five children are balancing?

  1. A21 minutes 30 seconds
  2. B22 minutes 15 seconds
  3. C22 minutes 30 seconds
  4. D23 minutes 15 seconds
  5. E23 minutes 45 seconds

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Lesson 5 of 7Patterns and SequencesIntermediate

The pattern goes 16, 20, 24, … — adding 4 each time. To count how many terms reach 404, use the nth-term formula: first + (n−1) × step = last. Solve for n.

Arithmetic sequence term-count questions appear regularly in the easy-to-medium band of Selective MR. Students who count up one-by-one from 16 will run out of time. The nth-term formula is the reliable shortcut.

The examiner tests whether students (1) identify the common difference (+4), (2) set up 16 + (n−1)×4 = 404, (3) correctly solve for n (n−1 = 97, so n = 98), and (4) avoid the off-by-one trap of answering 97 instead of 98.

An arithmetic sequence is defined by its first few terms and its last term. Students find how many terms are in the sequence.

Best approach: nth term = first + (n−1) × d. Substitute first = 16, d = 4, last = 404. Solve: (n−1) = (404−16)÷4 = 388÷4 = 97. So n = 97 + 1 = 98. Always add 1 at the end (counting fence posts, not gaps).

Question

The first three numbers in a number pattern are:

16, 20, 24, …

The last number is 404.

How many numbers are in this number pattern?

  1. A92
  2. B93
  3. C97
  4. D98
  5. E101

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Lesson 6 of 7Patterns and SequencesIntermediate

Two alternating rules: −100 then +10, repeated. Work backwards from the 6th term by applying the inverse of each rule in reverse order.

Alternating-rule sequences appear in the medium-to-difficult band of Selective MR. Students who only apply one rule (e.g. just add 90 each step) will choose a wrong starting value. The key is tracking which rule applies at each position.

The examiner tests whether students (1) correctly alternate the two rules (not apply them once each), (2) work backwards by applying the inverse operations in reverse order, and (3) verify by running the sequence forward to check.

A sequence uses two alternating rules. A term part-way through is given. Students find the first term by working backwards.

Best approach: Map out which rule applies at each step: step 1→2 is −96100, step 2→3 is +10, etc. Work backwards: undo the last step first (add 100 to undo −100), then undo the step before that (+100 or −10), and repeat until you reach position 1.

Question

Jamal makes a number sequence. He chooses the 1st number in the sequence.

Then he follows these two rules, one after the other, repeatedly:

  • Subtract 100 to get the next number in the sequence.
  • Add 10 to get the next number in the sequence.

The 6th number in the sequence is 8451.

What is the 1st number in Jamal's sequence?

  1. A8171
  2. B8181
  3. C8711
  4. D8721
  5. E8731

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Lesson 7 of 7Patterns and SequencesDifficult

Each number (from the 4th) = the three before it added together. Write equations for Y and the known term (6), solve for X, then find Z in one more step.

Tri-additive sequences with missing interior terms appear in the difficult band of Selective MR. Students who jump straight to Z without finding X and Y first always get stuck.

The examiner tests whether students can set up and solve two simultaneous equations using the tri-additive rule, working with both known and unknown positions in the sequence.

A sequence uses the rule ‘each term = sum of the three before it’. Some terms are given and some are labelled with letters. Students solve for the unknowns step by step.

Best approach: Write the rule for each unknown term as an equation. Use a known term to solve for X first, then substitute to find Y, then finally calculate Z.

Question

Claudia makes a number pattern using these rules:

  • Choose the first three numbers.
  • After this, each number is the sum of the three numbers before it.

Here is Claudia’s pattern with three missing numbers labelled X, Y and Z.

2, X, 0, Y, 6, Z

All the numbers are whole numbers.

What number does Z represent?

  1. A8
  2. B9
  3. C10
  4. D11
  5. E12

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