Connect arithmetic and geometric progressions in one problem
Use three-term conditions, retain possible ratio signs and choose separate formulae when a problem combines different progressions.
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
- Use AP and GP nth terms and finite sums.
- Solve quadratic equations and apply a supplied positivity condition.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which relation is required for consecutive AP terms a, b, c?
What you will learn
- Use the three-term AP and GP conditions with appropriate restrictions.
- Solve linked progression problems and compare totals using each sequence's own rule.
Translate three terms into a relation
Three consecutive AP terms a, b, c satisfy b − a = c − b, equivalently 2b = a + c. For non-zero consecutive GP terms, b/a = c/b, equivalently b² = ac. Check the original differences or ratios after solving. The squared relation alone needs care with zeros: 0, 0, 5 satisfies b² = ac but cannot be generated by a constant ratio from its first zero term.
If x, x + 3 and 2x + 6 are positive consecutive GP terms, (x + 3)² = x(2x + 6) gives x² = 9. The candidates are x = ±3; positivity selects x = 3 and terms 3, 6, 12 with ratio 2. Without the stated positivity condition, the candidate x = −3 yields −3, 0, 0, which is a GP under the ratio-zero recurrence and must be assessed rather than discarded automatically.
Do not carry one progression's rule into another
The AP 3, 5, 7, 9, … adds 2 each time and has first-four sum 24. The GP 3, 6, 12, 24, … multiplies by 2 and has first-four sum 45. Both start at 3, but their nth terms and total formulae differ. Constant non-zero progressions can be both APs with d = 0 and GPs with r = 1; labels depend on the defining relation, not on whether values happen to increase.
Indexed GP information can leave two ratio signs. If u₂ = 6 and u₄ = 24, then r² = 4, so r = 2 or −2. Recovering a = 6/r gives (a, r) = (3, 2) or (−3, −2). Both match the supplied positive even-indexed terms; they have different odd-indexed terms and different totals. Retain both unless another condition selects one.
Pause and explain
If a GP has u₂ = 6 and u₄ = 24, which ratios are possible?
Worked example
Positive numbers x, x + 3 and 2x + 6 are consecutive GP terms. Find x and the common ratio.
Show the worked solution
- Use (x + 3)² = x(2x + 6), giving x² + 6x + 9 = 2x² + 6x.
- Thus x² = 9, with candidates x = 3 and −3; the requirement that all terms are positive selects x = 3.
- The terms are 3, 6, 12, and both consecutive ratios are 2.
Answer x = 3; terms 3, 6, 12; ratio 2.
Try a topic test
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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.
Find k such that 2, k + 1 and 3k − 2 are consecutive AP terms.
Give me a hint
Set twice the middle term equal to the sum of its neighbours.
Compare my reasoning
- Write 2(k + 1) = 2 + (3k − 2).
- This gives 2k + 2 = 3k, hence k = 2.
- The terms become 2, 3, 4, with equal differences 1, confirming the condition.
k = 2; terms 2, 3, 4; common difference 1.
Look for these in your work
- Translate the AP condition into an equation.
- Verify both original consecutive differences.
A GP has u₂ = 6 and u₄ = 24. Find all (a, r) pairs and their respective S₅ values.
Give me a hint
Solve for both ratio signs before finding the first term and sum.
Compare my reasoning
- r² = 24/6 = 4, so r = 2 or −2; a = 6/r gives 3 or −3 respectively.
- For (3, 2), S₅ = 3(2⁵ − 1)/(2 − 1) = 93.
- For (−3, −2), the terms −3, 6, −12, 24, −48 total −33; both pairs match the given indexed terms.
(a, r) = (3, 2) gives S₅ = 93; (−3, −2) gives S₅ = −33.
Look for these in your work
- Retain both ratio signs from the even power.
- Compute and verify the total for each candidate progression.
Two supplied four-week practice-time models are A: 20, 25, 30, 35 minutes and B: 10, 20, 40, 80 minutes. Identify the progressions and compare their four-week totals.
Give me a hint
Use a constant difference for A and a constant ratio for B; compare total time over the same four weeks.
Compare my reasoning
- A is an AP with a = 20 and d = 5, so S₄ = 4(20 + 35)/2 = 110 minutes.
- B is a GP with a = 10 and r = 2, so S₄ = 10(2⁴ − 1)/(2 − 1) = 150 minutes.
- B's supplied model totals 40 minutes more. These are time totals, not evidence of relative learning outcomes or a justification for extending either model indefinitely.
A: AP, 110 min; B: GP, 150 min; B totals 40 min more.
Look for these in your work
- Recognise each progression from its own consecutive relation.
- Compare sums over the same indexed interval with appropriate units.
Common mistakes
- Check the original progression condition, especially when some terms are zero.
- Positive indexed observations do not necessarily force a positive ratio.
- Use a separate rule and sum formula for every progression in a combined problem.
What can you explain now?
Find all real p such that 3, p, 27 are consecutive GP terms.
Compare with the explanation
p = 9 or −9, with ratios 3 or −3 respectively.
The squared relation p² = 81 gives both signs; each yields equal consecutive non-zero ratios.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain when 2b = a + c or b² = ac is useful, then give a progression problem with two valid ratio signs and describe the extra condition that would select one.
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.