Find arithmetic progression terms and finite sums
Recognise a constant difference, recover the first term from indexed data and turn a total constraint into a valid integer term count.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use positive-integer binomial powers and positive-integer term indices. Written practice is self-checked.
Before you begin
- Solve simultaneous linear equations.
- Distinguish the first term, an indexed term and a cumulative total.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which progression is arithmetic?
What you will learn
- Recognise and construct an arithmetic progression using its common difference.
- Use the nth-term and first-n-term sum formulae, including practical constraints on n.
A constant difference determines the nth term
An arithmetic progression has the same difference d between consecutive terms. Starting with first term a, the nth term is uₙ = a + (n − 1)d for positive integers n. There are n − 1 steps from the first term to the nth. For 7, 10, 13, …, a = 7 and d = 3; the twelfth term is 7 + 11(3) = 40. Negative differences are allowed and produce decreasing progressions.
Given indexed terms, write separate equations using their actual indices. If u₄ = 14 and u₁₀ = 38, then a + 3d = 14 and a + 9d = 38. Subtraction gives 6d = 24, so d = 4 and a = 2. Thus uₙ = 4n − 2. The first supplied observation need not be the first term of the progression.
Pair opposite ends to obtain a sum
Pair the first and last terms, the second and next-to-last, and so on. Each pair has sum a + uₙ. This gives Sₙ = n(a + uₙ)/2 = n[2a + (n − 1)d]/2, with the same formula working for odd n by accounting for the middle term. Sₙ is a total, not the nth individual value. For a = 2 and d = 4, S₂₀ = 20(2 + 78)/2 = 800.
In a count problem n must be a positive integer, and a physical interpretation may further require positive terms. A seating model with first row 18 seats and two more seats per row has Sₙ = n(n + 17). Fifteen rows give 480 seats and sixteen give 528, so sixteen is the smallest count reaching at least 500. Check neighbouring integer totals rather than rounding an inequality solution without justification.
Pause and explain
An AP starts at 7 and increases by 3. Which expression gives u₁₂?
Worked example
An arithmetic progression has u₄ = 14 and u₁₀ = 38. Find its nth term and S₂₀.
Show the worked solution
- Write a + 3d = 14 and a + 9d = 38, giving d = 4 and a = 2.
- The nth term is uₙ = 2 + 4(n − 1) = 4n − 2, so u₂₀ = 78.
- Use S₂₀ = 20(2 + 78)/2 = 800 and verify both supplied indexed terms.
Answer uₙ = 4n − 2; S₂₀ = 800.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For 7, 10, 13, …, find the twelfth term and the sum of the first twelve terms.
Give me a hint
Find the last term first, then use the mean of the first and last terms for the total.
Compare my reasoning
- The common difference is 3, so u₁₂ = 7 + 11(3) = 40.
- The first-twelve sum is S₁₂ = 12(7 + 40)/2.
- This equals 282; the single term 40 is different from the cumulative sum.
u₁₂ = 40; S₁₂ = 282.
Look for these in your work
- Use eleven differences for the twelfth term.
- Use the actual first and last values to calculate the sum.
An AP has u₃ = 11 and u₈ = 31. Find a, d, uₙ and S₁₅.
Give me a hint
Subtract the indexed-term equations before recovering a.
Compare my reasoning
- a + 2d = 11 and a + 7d = 31 give 5d = 20, hence d = 4 and a = 3.
- Then uₙ = 3 + 4(n − 1) = 4n − 1 and u₁₅ = 59.
- The sum is S₁₅ = 15(3 + 59)/2 = 465; check u₃ and u₈ in the rule.
a = 3; d = 4; uₙ = 4n − 1; S₁₅ = 465.
Look for these in your work
- Respect the supplied term indices.
- Verify the recovered rule and calculate a total rather than just its last term.
A hall has 18 seats in row one and two more in every subsequent row. Under this model, find the smallest number of rows providing at least 500 seats.
Give me a hint
Use the AP total, then compare adjacent positive integer counts.
Compare my reasoning
- With a = 18 and d = 2, Sₙ = n[36 + 2(n − 1)]/2 = n(n + 17).
- S₁₅ = 15(32) = 480, below the target, while S₁₆ = 16(33) = 528 reaches it.
- Each added row has a positive number of seats, so these increasing totals establish that 16 is the smallest valid count.
16 rows, providing 528 seats; 15 rows provide only 480.
Look for these in your work
- Model the capacity as a sum of rows.
- Justify the minimum with adjacent integer totals and increasing capacity.
Common mistakes
- Use n − 1 differences, not n, when counting from the first term.
- An indexed observation is not automatically the first term.
- Check integer counts and the model's permitted domain in total problems.
What can you explain now?
An AP has a = 7 and d = 3. Find u₁₂ and S₁₂.
Compare with the explanation
u₁₂ = 40; S₁₂ = 282.
There are eleven steps to term twelve, and its twelve-term total is 12(7 + 40)/2.
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Make it stick
Derive uₙ and Sₙ from the meaning of a common difference, then explain why a minimum row count needs an integer and a neighbouring check.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.