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Learn7357Paper 1 · PureSequences and seriesGeometric progressions
Sequences and series · Subtopic 3 of 5

Geometric Progressions

Sequences that multiply by the same number each time: the nnth term, the sum of the first nn terms and its proof, finding unknowns, and the growth and decay problems they model.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Recognise a geometric progression (GP) and find its common ratio rr.
  • Use un=arn−1u_n=ar^{n-1} and Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}.
  • Find aa, rr or nn from given information.
  • Model compound growth and decay with GPs.
Pause and think

Fold a sheet of paper 0.10.1 mm thick in half, then in half again, and so on. How thick would it be after 4242 folds?

Section 2 of 14

Core idea

Start hereRice on a chessboard

A legend: a king offered a reward, and the inventor of chess asked for 11 grain of rice on the first square, 22 on the second, 44 on the third, doubling each time. It sounds modest, but the total is 264−1≈1.8×10192^{64}-1\approx1.8\times10^{19} grains, more rice than has ever been grown. Multiplying by the same number each step is a geometric progression, and its sum has a neat formula.

The big idea

A geometric progression (GP) multiplies by a common ratio rr each time. un=arn−1u_n=ar^{n-1} and Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r} (or a(rn−1)r−1\dfrac{a(r^n-1)}{r-1}), for r≠1r\neq1.

1. The nth term

Growth by a constant ratio
1234567204060801000term number nterm
Each bar is twice the one before. With r>1r>1 a GP grows faster and faster (exponential growth); with 0<r<10<r<1 it shrinks towards zero.
un=arn−1,r=un+1unu_n=ar^{n-1},\qquad r=\frac{u_{n+1}}{u_n}
where
aa
first term.
rr
common ratio: any term divided by the one before.
Worked exampleFrom two terms

A GP has 2nd term 1212 and 5th term 324324. Find aa and rr.

Goal
aa, rr.
Strategy
Divide the equations to eliminate aa.
  1. ar=12,ar4=324ar=12,\qquad ar^4=324
    u2u_2, u5u_5.
  2. ar4ar=r3=27⇒r=3\frac{ar^4}{ar}=r^3=27\Rightarrow r=3
    Divide.
  3. a=4a=4
    4×3=124\times3=12.
Answera=4a=4, r=3r=3.
✓ Check4,12,36,108,3244,12,36,108,324 ✓.

2. The sum: multiply and subtract

Proof of Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}
  1. Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}
    Definition.
  2. rSn=ar+ar2+⋯+arn−1+arnrS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n
    Multiply every term by rr: everything shifts one place.
  3. Sn−rSn=a−arnS_n-rS_n=a-ar^n
    Subtract: all the middle terms cancel.
  4. Sn(1−r)=a(1−rn)S_n(1-r)=a(1-r^n)
    Factorise both sides.
  5. Sn=a(1−rn)1−r(r≠1)S_n=\frac{a(1-r^n)}{1-r}\quad(r\neq1)
    Divide by 1−r1-r.
Worked exampleThe chessboard total

Find 1+2+4+⋯1+2+4+\cdots to 6464 terms.

Goal
The sum.
Strategy
a=1a=1, r=2r=2; use the r>1r>1 form.
  1. S64=1(264−1)2−1=264−1S_{64}=\frac{1(2^{64}-1)}{2-1}=2^{64}-1
    Formula.
  2. ≈1.84×1019\approx1.84\times10^{19}
    Calculator.
Answer264−1≈1.84×10192^{64}-1\approx1.84\times10^{19} grains.
✓ CheckSmall case: 3 squares, 1+2+4=7=23−11+2+4=7=2^3-1 ✓.
Worked exampleSmallest n with a sum over 5000

For 3+6+12+⋯3+6+12+\cdots, find the least nn with Sn>5000S_n>5000.

Goal
Whole nn.
Strategy
Inequality; take logs to bring nn down.
  1. 3(2n−1)>5000⇒2n>1667.73(2^n-1)>5000\Rightarrow2^n>1667.7
    Sum formula.
  2. nln⁡2>ln⁡1667.7⇒n>10.7n\ln2>\ln1667.7\Rightarrow n>10.7
    Take logs; ln⁡2>0\ln2>0 so no flip.
Answern=11n=11.
✓ CheckS10=3069S_{10}=3069, S11=6141S_{11}=6141 ✓.
Edge caseNegative ratio 1−2+4−8+⋯1-2+4-8+\cdots has r=−2r=-2: terms alternate in sign. The formula still works: S8=1(1−256)3=−85S_8=\frac{1(1-256)}{3}=-85.
TrickAP or GP? Subtract consecutive terms: constant ⇒ AP. Divide: constant ⇒ GP.
Pause and think

A salary of 30 000 rises 4% a year. Total over 10 years?

Summary

  • un=arn−1u_n=ar^{n-1}.
  • Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} from multiply-and-subtract.
  • Logs to find nn.
Key words
Common ratio
The constant multiplier.
HistoryGrains on a chessboard. An old legend tells of a ruler who offered the inventor of chess any reward. He asked for one grain of rice on the first square, two on the second, four on the third, doubling each time. The total is 264−1≈1.8×10192^{64}-1\approx1.8\times10^{19} grains, far more rice than has ever been grown on Earth.
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