Common ratio, nth term, sum to n terms, sum to infinity, convergence.
Answer at least 3 of 3 correctly to complete this section.
Geometric sequences model any situation where a quantity is repeatedly multiplied by the same factor: compound interest, radioactive decay, population growth, the bounce height of a ball. The sum to infinity formula is one of the most powerful results in A-Level maths — it lets you find a finite total from infinitely many terms, which is the gateway to understanding convergence.
A geometric sequence is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio, r.
2,6,18,54,162,…r=3
100,50,25,12.5,…r=0.5
1,−2,4,−8,16,…r=−2
To check whether a sequence is geometric, divide any term by the previous term. If the ratio is the same every time, the sequence is geometric:
r=unun+1
If the first term is a and the common ratio is r, then:
un=arn−1
Why rn−1 and not rn? Because to get from the first term to the nth term, you multiply by r exactly n−1 times. The first term is ar0=a.
For the sequence 2,6,18,54,… we have a=2 and r=3, so:
un=2×3n−1
Check: u1=2×30=2 ✓, u4=2×33=54 ✓
Drag the sliders to build different geometric sequences. Watch how changing r affects whether the terms grow, shrink, or alternate:
Adjust the first term and common ratio to explore geometric sequences
Common mistake: Students write un=arn instead of un=arn−1. With a=2,r=3: the correct formula gives u1=2×30=2, but arn=2×31=6. The power is always one less than the term number.
Answer at least 3 of 4 correctly to complete this section.
A geometric series is what you get when you add up the terms of a geometric sequence:
Sn=a+ar+ar2+ar3+…+arn−1
There is a neat algebraic trick. Write out Sn and then rSn:
Sn=a+ar+ar2+…+arn−1
rSn=ar+ar2+ar3+…+arn
Subtract the second from the first — almost all the terms cancel:
Sn−rSn=a−arn
Sn(1−r)=a(1−rn)
Sn=1−ra(1−rn)(r=1)
When r=1, every term equals a, so Sn=na.
Find the sum of the first 6 terms of the geometric series with a=3 and r=2.
S6=1−23(1−26)=−13(1−64)=−13×(−63)=189
Check: 3+6+12+24+48+96=189 ✓
A geometric series has a=5, r=2, and Sn=635. Find n.
1−25(1−2n)=635
−15(1−2n)=635
5(2n−1)=635
2n−1=127
2n=128=27
So n=7.
Answer at least 3 of 4 correctly to complete this section.
Consider the geometric series with a=1 and r=21:
1+21+41+81+161+…
Even though there are infinitely many terms, the sum approaches 2. Each term gets smaller and smaller, contributing less and less to the total. We say the series converges.
This happens whenever ∣r∣<1 — each term is smaller than the last (in absolute value), so the terms shrink towards zero.
A geometric series converges if and only if:
∣r∣<1
This means −1<r<1. It does not matter whether r is positive or negative — what matters is its size.
When ∣r∣<1, we can take the limit of Sn as n→∞. Since ∣r∣<1, the term rn→0:
S∞=limn→∞1−ra(1−rn)=1−ra(1−0)=1−ra
Find the sum to infinity of the series 12+6+3+1.5+…
Here a=12 and r=126=21. Since ∣r∣=21<1, the series converges.
S∞=1−2112=2112=24
A geometric series has sum to infinity 20 and first term 8. Find r.
S∞=1−ra=20
1−r8=20
8=20(1−r)
8=20−20r
20r=12
r=2012=53
Check: ∣r∣=53<1 ✓, so the series does converge.
The sum to infinity formula can prove that recurring decimals are fractions. For example:
0.3=0.3+0.03+0.003+…
This is a geometric series with a=0.3 and r=0.1:
S∞=1−0.10.3=0.90.3=31
Exam tip: Always state the convergence condition. Write “Since ∣r∣=21<1, the series converges, so S∞ exists.” Examiners give marks for this.
Answer at least 3 of 4 correctly to complete this section.
These questions combine multiple skills and are typical of what you will see on an A-Level paper.
The 2nd term of a geometric sequence is 6 and the 5th term is 162. Find a and r.
u2=ar=6...(1)
u5=ar4=162...(2)
Divide (2) by (1):
arar4=6162
r3=27
r=3
Substitute back into (1): a×3=6, so a=2.
A geometric series has first term a and common ratio r. The sum to infinity is 4 times the first term. Find r.
S∞=4a
1−ra=4a
1−r1=4
1=4(1−r)
1=4−4r
4r=3
r=43
A ball is dropped from a height of 10 metres. After each bounce it reaches 53 of its previous height. Find the total distance the ball travels.
The ball falls 10 m, then bounces up 6 m, falls 6 m, bounces up 3.6 m, falls 3.6 m, and so on.
Total distance = initial drop + 2 × (sum of bounce heights)
The bounce heights form a geometric series: 6+3.6+2.16+… with a=6, r=53.
S∞=1−536=526=15
Total distance =10+2×15=40 metres.
Show that S∞−Sn=1−rarn for a convergent geometric series.
S∞−Sn=1−ra−1−ra(1−rn)=1−ra−a(1−rn)=1−ra−a+arn=1−rarn□
This tells you the “error” when you approximate the infinite sum by the first n terms.
Answer at least 3 of 4 correctly to complete this section.
| Formula | When to use |
|---|---|
| un=arn−1 | Find a specific term |
| r=unun+1 | Find the common ratio |
| Sn=1−ra(1−rn) | Sum of first n terms (r=1) |
| S∞=1−ra | Sum to infinity (∥r∥<1) |
| Situation | Method |
|---|---|
| Find r from two terms | Divide: rk=um−kum, then take the kth root |
| Is the series convergent? | Check whether ∥r∥<1 |
| Find n given Sn | Substitute into sum formula and solve (often using logarithms) |
| Find a and r | Use two given facts to set up and divide equations |
| Recurring decimal as fraction | Write as geometric series with r=0.1 or 0.01 etc. |
Lock in what you've learned with exam-style questions and spaced repetition.