Algebra - Sequences: deduce expressions to calculate the nth term of linear sequences (Foundation tier)

Overview

A linear sequence is a list of numbers that changes by the same amount at every step. Your job is to turn that repeated change into an expression that finds any term directly. Instead of adding 3 nineteen times to reach the twentieth term, you can substitute 20 into a short algebraic rule. This is Pearson Edexcel GCSE Mathematics 1MA1, Foundation tier, specification A25: deduce expressions to calculate the nth term of linear sequences. 1

Candidates may be asked to find an expression in terms of n, calculate a distant term, explain a mistake, or decide whether a number appears in a sequence. These questions connect number patterns with substitution, solving equations and straight-line graphs. An nth-term expression is often a short question, but linked subparts can test several skills together.

The central idea is Difference, Adjust, Check. Find the common difference, write that number multiplied by n, then adjust by a constant. Finally, test a later given term. Throughout this guide, n starts at 1 unless the question explicitly says otherwise. Every question created here is original Edexcel-style practice; its stated marks are teaching allocations, not an official Pearson mark scheme.

Key Concepts

1. Position and value are different

In the sequence 5, 8, 11, 14, the first term has value 5 and the fourth term has value 14. The position is a counting number: 1, 2, 3, 4 and so on. The term value is the number occupying that position. Write u_n for the value at position n; this notation means the nth term, not u multiplied by n.

The expression 3n + 2 is a position-to-term rule. Substituting n = 4 gives 3 × 4 + 2 = 14. By contrast, “add 3” is a term-to-term rule: it tells you how to move from one value to the next. Both describe the pattern, but only the expression answers “Find the nth term in terms of n”.

Cover and recall: In 5, 8, 11, 14, what are the position and value of the third term? Say “position 3, value 11” before checking. Candidates who confuse these two quantities often substitute the term value where the position belongs.

2. Identify the common difference, including its sign

Subtract each term from the next. For 5, 8, 11, 14, the differences are 8 − 5 = 3, 11 − 8 = 3 and 14 − 11 = 3. Equal differences identify an arithmetic, or linear, sequence in this GCSE context. Increasing sequences have positive differences; decreasing sequences have negative differences.

For 20, 17, 14, 11, the difference is 17 − 20 = −3. “The gap is 3” is incomplete if it hides the direction. Write −3 beside each gap. The coefficient of n must therefore be −3, not +3.

Check every available gap. The list 2, 4, 7, 11 has differences 2, 3, 4, so the linear method does not apply. Do not invent a constant difference from only the first pair. A few equal gaps do not prove that every possible continuation is linear, but a question stating “arithmetic sequence” supplies the intended pattern.

3. Why the difference is the coefficient of n

The multiples of 3 are 3, 6, 9, 12, generated by 3n. When n increases by 1, 3n increases by 3. Adding a fixed constant changes each value equally, so it does not change the gaps. Therefore 3n + 2 still has common difference 3.

Algebra confirms the idea. Consecutive terms of dn + c are dn + c and d(n + 1) + c. Subtracting gives d(n + 1) + c − (dn + c) = d. The constant cancels, leaving the coefficient d. You do not need to reproduce this proof in every exam answer, but understanding it prevents guesswork.

Why does this work? If you add 10 to every term, the sequence moves upwards by 10 but neighbouring terms remain the same distance apart. Only the constant changes. If you change the coefficient instead, the gaps change, so you have a different rate of growth.

4. Find the constant by matching positions

For 5, 8, 11, 14, begin with 3n. At positions 1, 2, 3, 4, that produces 3, 6, 9, 12. Each actual value is 2 larger than its matching multiple. Add 2 to obtain 3n + 2.

The quick version is c = first term − difference. Here c = 5 − 3 = 2. The first term is not automatically the constant. Writing 3n + 5 would give 8 at position 1, which is already wrong.

For 3, 7, 11, 15, the difference is 4. The multiples 4, 8, 12, 16 are each 1 too large. Subtract 1, giving 4n − 1. A negative constant is perfectly valid. In 6, 12, 18, 24, the constant is 6 − 6 = 0, so the answer simplifies to 6n.

Use the diagram actively: Cover the last two boxes. For 7, 12, 17, 22, work out the coefficient and constant yourself. The completed expression is 5n + 2. Explain why the constant is 2 rather than 7.

5. Decreasing sequences and equivalent expressions

For 20, 17, 14, 11, start with −3n. At position 1, −3n gives −3, so add 23 to reach 20. The expression is −3n + 23, usually written 23 − 3n. These forms are equivalent and generate the same terms.

The shortcut still works: c = 20 − (−3) = 23. Subtracting a negative adds its positive counterpart. Keep the brackets until you have handled the signs correctly.

Check position 4: 23 − 3 × 4 = 11. The incorrect rule 23 + 3n produces 35 instead. A decreasing sequence can continue through zero into negative term values. For this sequence, the eighth term is −1. That is allowed because the position 8 is positive, even though the term value is negative.

6. Calculate a distant term accurately

Once you know the expression, substitute the requested position. For u_n = 3n + 2, the twentieth term is u_20 = 3 × 20 + 2 = 62. Multiplication comes before addition. The expression means “multiply n by 3, then add 2”, not “multiply 3 by n + 2”.

Use n = 20, not n = 19. There are nineteen jumps from the first term to the twentieth, but the rule 3n + 2 has already accounted for the starting position. If using the alternative first-term formula, those nineteen jumps appear explicitly.

Candidates should show the substitution, not merely an unexplained number, particularly when the question requests working. A sequence question can appear on any of the three papers. Practise this method without a calculator so it remains available on Paper 1. 1

7. Decide whether a number belongs

Set the nth-term expression equal to the target value and solve for n. To test 68 in 3n + 2, write 3n + 2 = 68, then 3n = 66 and n = 22. Since 22 is a positive whole number, 68 is the twenty-second term. Check: 3 × 22 + 2 = 68.

To test 69, solve 3n + 2 = 69. This gives n = 67/3, which is not a whole number. Therefore 69 is not a term. Do not round to 22: rounding would answer a different question.

A whole-number solution is not enough if it is zero or negative. Testing 2 gives n = 0, which is not an allowed position here. Testing −1 gives n = −1, also disallowed. The decisive condition is positive whole-number position, not positive term value.

For a finite real-world pattern, check its stated limit too. If only twenty rows are built, an answer n = 22 is outside that context. If d = 0, the sequence is constant; compare the target with that constant instead of dividing by zero.

Mathematical/Scientific Relationships

The relationships to learn

All the sequence relationships below are labelled Must memorise because they are not supplied on the checked 2026 Edexcel Foundation formula sheet. Understanding how to reconstruct them is just as useful as recalling them. 3

Must memorise: d = u_(n+1) − u_n. Here d is the common difference, u_n is a term value, and u_(n+1) is the next value. Use subtraction in this order to preserve the sign.

Must memorise: u_n = dn + c, with c = a − d. Here a is the first term, n is a positive whole-number position, and c is the fixed adjustment. This is the main Foundation method.

Must memorise: u_n = a + (n − 1)d. This is an equivalent form. There are n − 1 jumps after the first term. For a = 5 and d = 3, it becomes 5 + 3(n − 1) = 3n + 2. Learn one method securely rather than switching mid-solution.

Must memorise: u_(n+1) = u_n + d. This is the term-to-term relationship. It generates the next value but is not the requested nth-term expression.

Must memorise: n = (T − c)/d when d ≠ 0. Here T is the target term value. This rearrangement tests membership; interpret the result as a position.

Must memorise: d = (u_q − u_p)/(q − p), where q ≠ p. This finds the difference when two known positions are not adjacent. If the third term is 11 and the seventh is 23, four position steps produce an increase of 12, so d = 3. Then c = 11 − 3 × 3 = 2.

Graph and data skills

Plot position n horizontally and term value u_n vertically. For 3n + 2, plot (1, 5), (2, 8), (3, 11), (4, 14). Use equal scale intervals, label both axes and keep coordinates in position-first order. The points lie on a straight line with gradient 3.

The extended line crosses the vertical axis at 2, matching c, but n = 0 is not a term position in this sequence. The sequence consists of points at positive integer positions, not every point on the continuous line. Use a line extension for interpretation only when appropriate to the question; do not claim that position 2.5 is allowed.

Practical Applications

Growing arrangements

A seating pattern might have 5 seats in row 1, 8 in row 2, 11 in row 3 and 14 in row 4. If the pattern continues, row n has 3n + 2 seats. The thirtieth row therefore has 92 seats. This is the number in that row, not the total number of seats across thirty rows. Read the quantity being requested before calculating.

Money and units

A collection starts with £5 in week 1 and increases by £3 each week. Its value in week n is £(3n + 2). If the question uses pence, convert consistently: the rule becomes 300n + 200 pence because £1 = 100p. The position n has no money units. There is no required practical for this algebra topic.

Exam-Focused Checking and Recall

A real marking example matters more than a guessed universal rule. In Pearson's June 2024 Paper 1F mark scheme, question 20 awards B2 for 4n − 3 or equivalent. It allows B1 for an unambiguous 4n + k with the constant wrong or absent. This demonstrates why writing the coefficient with n can secure partial credit in that question. Other questions have their own schemes; a standalone difference written alone is not guaranteed a mark. 2

For “Find an expression”, finish with an expression containing n. For “Work out the twentieth term”, finish with a value. For “Explain whether”, show a calculation and an interpreted conclusion. Writing u_n = 3n + 2 is legitimate notation; writing n = 3n + 2 confuses position with value.

Test the last given term, since it checks both the difference and the adjustment. A rule can accidentally fit the first term while using the wrong coefficient. Avoid presenting several competing final answers. Cross out an abandoned expression clearly.

Recall routine: Close the guide and find the rule for 9, 13, 17, 21. Next, explain why 41 is a term but 42 is not. Check afterwards: the rule is 4n + 5; 41 gives n = 9, while 42 gives n = 37/4. Retry a decreasing sequence tomorrow, then mix both types a few days later.

Ten-Minute Tutor Podcast

Listen actively: introduction 0:00–1:00, core concepts 1:00–6:00, exam technique 6:00–8:00, recall quiz 8:00–9:00, summary 9:00–10:00. Say answers aloud during the quiz and pause the player if you need more thinking time.

Tier boundary: Everything taught here is Foundation-accessible. Deduction of quadratic nth-term expressions is Higher-only in 1MA1 and is outside this guide. Recognising that unequal first differences are not linear remains useful at Foundation. 1

References

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