Find geometric progression terms and finite sums
Recognise a constant multiplier, recover the term rule and sum a finite progression without assuming the ratio is positive or less than one.
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 positive integer powers and preserve signs.
- Rearrange equations and distinguish a term from a sum.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the common ratio of 12, −6, 3, …?
What you will learn
- Recognise a geometric progression and find its common ratio.
- Use the nth-term and finite-sum formulae, including negative and unit ratios.
Each next term is multiplied by the same ratio
A geometric progression starts with a and obeys uₙ₊₁ = ruₙ for a constant ratio r. For non-zero consecutive terms, divide the later by the earlier to find r. The nth term is uₙ = ar^(n − 1). For 3, 6, 12, …, r = 2; for 12, −6, 3, …, r = −1/2 and the signs alternate. Equal first terms alone do not determine whether a sequence is arithmetic or geometric.
With u₂ = 6 and u₅ = 48, divide their expressions ar and ar⁴ to get r³ = 8, hence r = 2 and a = 3. The gap between indices is three, so it gives r³ rather than r⁵. A ratio of zero generates a, 0, 0, …; use the recurrence in that case instead of dividing 0 by 0. The first term is a, and later terms are zero.
Subtract a shifted sum to derive the finite formula
Write Sₙ = a + ar + … + ar^(n − 1). Multiplying by r shifts every term, so Sₙ − rSₙ = a − arⁿ. For r ≠ 1 this yields Sₙ = a(1 − rⁿ)/(1 − r), equivalently a(rⁿ − 1)/(r − 1). These finite formulae work for growing and alternating progressions as well as shrinking ones. For r = 1, every term is a and Sₙ = na; the quotient formula would divide by zero.
For a = 3 and r = 2, S₈ = 3(2⁸ − 1)/(2 − 1) = 765. For a = 12 and r = −1/2, S₆ = 12[1 − (−1/2)⁶]/[1 − (−1/2)] = 63/8. Keep the sign inside the power and the denominator bracket. A finite total is always calculated over a stated number of terms; an infinite sum needs an additional convergence condition.
Pause and explain
Which formula gives the first n terms of a GP when r = 1?
Worked example
A GP has u₂ = 6 and u₅ = 48. Find a, r, uₙ and S₈.
Show the worked solution
- Divide u₅ by u₂ to obtain r³ = 48/6 = 8, so r = 2.
- From ar = 6, a = 3 and uₙ = 3 × 2^(n − 1).
- Use the finite sum S₈ = 3(2⁸ − 1)/(2 − 1) = 765 and check both indexed terms.
Answer a = 3; r = 2; uₙ = 3 × 2^(n − 1); S₈ = 765.
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 a GP with a = 5 and r = 3, find u₅ and S₅.
Give me a hint
Use four ratio steps for the fifth term, but five terms in the sum formula.
Compare my reasoning
- u₅ = 5 × 3⁴ = 405.
- S₅ = 5(3⁵ − 1)/(3 − 1) = 5(242)/2 = 605.
- The terms 5, 15, 45, 135, 405 total 605, independently checking the formula.
u₅ = 405; S₅ = 605.
Look for these in your work
- Distinguish the exponents for a term and a sum.
- Verify the total with the generated finite sequence.
Find the sixth term and first-six sum of 12, −6, 3, … exactly.
Give me a hint
Use r = −1/2 with brackets around its powers.
Compare my reasoning
- u₆ = 12(−1/2)⁵ = −3/8.
- S₆ = 12[1 − (−1/2)⁶]/[1 + 1/2] = 63/8.
- The six terms are 12, −6, 3, −3/2, 3/4, −3/8, whose sum is 63/8.
u₆ = −3/8; S₆ = 63/8.
Look for these in your work
- Retain the negative ratio and the correct exponent parity.
- Simplify the finite total exactly.
A 16 m spool supplies six pieces: the first is 8 m and each next piece is half the preceding one. Find the sixth length, total removed and amount remaining.
Give me a hint
Model the piece lengths as a finite GP, not as six equal cuts.
Compare my reasoning
- The piece GP has a = 8, r = 1/2, so u₆ = 8(1/2)⁵ = 1/4 m.
- The total is S₆ = 8[1 − (1/2)⁶]/(1/2) = 63/4 m.
- The remaining spool length is 16 − 63/4 = 1/4 m; six cuts do not remove the infinite limiting total.
Sixth piece 1/4 m; removed 63/4 m; remaining 1/4 m.
Look for these in your work
- Count six terms with the correct fifth ratio power for the last one.
- Subtract the finite removed total from the supplied stock length.
Common mistakes
- The ratio power depends on the difference between the supplied indices.
- A negative common ratio alternates signs; keep it inside the power bracket.
- Use Sₙ = na when r = 1 rather than dividing by zero.
What can you explain now?
A GP has first term 7 and ratio 1. Find u₁₀ and S₁₀.
Compare with the explanation
u₁₀ = 7; S₁₀ = 70.
Every term equals 7, so the ten-term total is ten copies of 7; the usual quotient formula is undefined here.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Derive the finite GP sum by subtracting rSₙ, and explain how to handle r = 1, r = 0 and a negative ratio without misusing a formula.
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.