nth term, common difference, sum of arithmetic series, applications.
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Arithmetic sequences are one of the simplest patterns in maths, but they model surprisingly many real situations: saving a fixed amount each month, seats in a cinema that increase row by row, or a runner who increases their distance by the same amount each week. The sum formula lets you add up hundreds of terms instantly — no calculator loop required.
An arithmetic sequence is a sequence where the difference between consecutive terms is always the same. That constant gap is called the common difference, d.
3,7,11,15,19,…d=4
20,17,14,11,8,…d=−3
To check whether a sequence is arithmetic, compute the difference between any two consecutive terms. If it is the same every time, the sequence is arithmetic.
If the first term is a and the common difference is d, then:
un=a+(n−1)d
Why (n−1) and not n? Because to get from the first term to the nth term, you add d exactly n−1 times. Think of it as starting at a and taking n−1 steps of size d.
For the sequence 3,7,11,15,… we have a=3 and d=4, so:
un=3+(n−1)×4=3+4n−4=4n−1
Check: u1=4(1)−1=3 ✓, u5=4(5)−1=19 ✓
Drag the sliders to build different arithmetic sequences and see how changing a and d affects the pattern:
Adjust the first term and common difference to explore arithmetic sequences
Common mistake: Students write un=an+d instead of un=a+(n−1)d. These are different! With a=3,d=4: the correct formula gives u1=3, but an+d=3(1)+4=7. Always use the formula from the formula book.
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Exam questions rarely give you a and d directly. Instead, you are given two facts and must set up equations.
The 5th term of an arithmetic sequence is 19 and the common difference is 4. Find the first term and the 20th term.
Using un=a+(n−1)d:
u5=a+4d=19
a+4(4)=19⟹a=19−16=3
Now find u20:
u20=3+19×4=3+76=79
The 3rd term of an arithmetic sequence is 8 and the 8th term is 23. Find a and d.
Write two equations:
u3=a+2d=8...(1)
u8=a+7d=23...(2)
Subtract (1) from (2): 5d=15, so d=3.
Substitute back: a+6=8, so a=2.
Is 150 a term of the arithmetic sequence 7,11,15,19,…?
Here a=7 and d=4. Set un=150:
7+(n−1)×4=150
4(n−1)=143
n−1=35.75
Since n must be a positive integer and 35.75 is not an integer, 150 is not a term of this sequence.
Exam tip: If you get a non-integer value for n, write a clear conclusion: “Since n is not a positive integer, the value is not a term in the sequence.” Examiners want to see this reasoning.
How many terms are there in the arithmetic sequence 5,9,13,…,201?
Set un=201 with a=5, d=4:
5+(n−1)×4=201
4(n−1)=196
n−1=49
n=50
There are 50 terms in the sequence.
Answer at least 3 of 4 correctly to complete this section.
A series is what you get when you add up the terms of a sequence. The sum of the first n terms of an arithmetic sequence is written Sn.
Sn=u1+u2+u3+…+un
The story goes that when Carl Friedrich Gauss was a schoolboy, his teacher asked the class to add up the numbers from 1 to 100. While the other students began adding one by one, Gauss found the answer almost immediately: 5050.
His insight: pair the first and last, second and second-to-last, and so on.
1+100=101 2+99=101 3+98=101 ⋮ 50+51=101
There are 50 such pairs, each summing to 101. So:
S100=50×101=5050
See this in action:
See why pairing first + last gives you the sum formula
Sequence:
Predict: What do you think the first term plus the last term equals? Make your prediction, then click to check.
This pairing trick works for any arithmetic series. If the first term is a and the last term is l:
Sn=2n(a+l)
Since l=a+(n−1)d, we can substitute to get the other form:
Sn=2n(2a+(n−1)d)
When to use which:
Find the sum of the first 20 terms of the arithmetic sequence 4,7,10,13,…
Here a=4, d=3, n=20.
S20=220(2×4+19×3)=10(8+57)=10×65=650
Find the sum of the arithmetic series 5+9+13+…+201.
First, find how many terms: 5+(n−1)×4=201⟹n=50.
Since we know the last term:
S50=250(5+201)=25×206=5150
The sum of the first n terms of the series 3+5+7+… is 168. Find n.
Here a=3, d=2. Using Sn=2n(2a+(n−1)d):
2n(6+2(n−1))=168
2n(6+2n−2)=168
2n(2n+4)=168
n(n+2)=168
n2+2n−168=0
(n+14)(n−12)=0
Since n must be positive, n=12.
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.
An arithmetic series has first term a and common difference d. The 10th term is 28 and S15=345. Find a and d.
From u10=a+9d=28...(1)
From S15=215(2a+14d)=345⟹2a+14d=46⟹a+7d=23...(2)
Subtract (2) from (1): 2d=5, so d=2.5.
Substitute back: a+7(2.5)=23⟹a=23−17.5=5.5.
Check: u10=5.5+9(2.5)=5.5+22.5=28 ✓
S15=215(2(5.5)+14(2.5))=215(11+35)=215×46=345 ✓
A runner trains by running 3 km on the first day, and increases the distance by 0.5 km each day. How far does the runner run in total over 30 days?
This is an arithmetic series with a=3, d=0.5, n=30.
S30=230(2×3+29×0.5)=15(6+14.5)=15×20.5=307.5 km
Show that the sum of the first n positive odd numbers is n2.
The odd numbers form the arithmetic sequence 1,3,5,7,… with a=1 and d=2.
Sn=2n(2(1)+(n−1)(2))=2n(2+2n−2)=2n(2n)=n2□
This is a classic result that examiners love to test.
Answer at least 3 of 4 correctly to complete this section.
| Formula | When to use |
|---|---|
| un=a+(n−1)d | Find a specific term |
| Sn=2n(2a+(n−1)d) | Find the sum when you know a and d |
| Sn=2n(a+l) | Find the sum when you know the last term l |
| Situation | Method |
|---|---|
| Find d from two terms | d=m−kum−uk |
| Is a value in the sequence? | Solve un=value; check n is a positive integer |
| Find n given Sn | Set up and solve a quadratic in n |
| Find a and d | Use two given facts to set up simultaneous equations |
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