Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Arithmetic progressions

Pure content (sections A–J) · Subtopic 19 of 66

Arithmetic Progressions

Sequences that go up (or down) by the same amount each time: the $n$ th term, the sum of the first $n$ terms and why it works, and how to find unknowns from given terms or sums.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Recognise an arithmetic progression (AP) and find its common difference $d$ .
Use $u_n=a+(n-1)d$ and $S_n=\tfrac n2\left[2a+(n-1)d\right]=\tfrac n2(a+l)$ .
Form and solve equations to find $a$ , $d$ or $n$ .
Use APs in real situations such as savings and seating.

Pause and think

Add up $1+2+3+\dots+100$ without adding one number at a time. Hint: pair the first number with the last.

Section 2 of 14

Core idea

Start hereThe schoolboy who added 1 to 100 in seconds

The story goes that a teacher set the class $1+2+3+\cdots+100$ to keep them busy, and young Carl Gauss answered $5050$ almost immediately. His trick: pair the first and last ( $1+100$ ), second and second-last ( $2+99$ ), and so on. Every pair makes $101$ , and there are $50$ pairs. That trick works for any sequence that goes up in equal steps.

The big idea

An arithmetic progression (AP) adds the same common difference $d$ each time. Its $n$ th term is $u_n=a+(n-1)d$ and the sum of $n$ terms is $S_n=\frac n2(a+l)=\frac n2[2a+(n-1)d]$ .

1. The nth term

An AP grows in equal steps

The tops of the bars lie on a straight line with gradient

d=4

. That is why APs are sometimes called linear sequences.

u_n=a+(n-1)d

$a$: the first term.
$d$: the common difference (any term minus the one before).
$n$: the position number.

Why $(n-1)$ ? To get from term $1$ to term $n$ you take $n-1$ steps of size $d$ .

Worked exampleFind $a$ and $d$

An AP has 3rd term $11$ and 8th term $31$ . Find $a$ , $d$ , and the 20th term.

Goal: $a$ , $d$ , $u_{20}$ .
Strategy: Two facts give two simultaneous equations.

$a+2d=11,\qquad a+7d=31$
$u_3$ and $u_8$ .
$5d=20\Rightarrow d=4$
Subtract the first from the second.
$a=11-8=3$
Back-substitute.
$u_{20}=3+19(4)=79$
$n$ th term formula.

Answer

a=3

d=4

u_{20}=79

✓ CheckSequence

3,7,11,15,19,23,27,31

: 3rd is

11

, 8th is

31

✓.

2. The sum: Gauss's trick

Two copies of a sum make a rectangle

Two copies of

1+2+3+4+5

fill a

5\times6

rectangle, so the sum is

\tfrac12\times5\times6=15

. In general

S_n=\tfrac n2(\text{first}+\text{last})

Proof of

S_n=\frac n2(a+l)

$S_n=a+(a+d)+(a+2d)+\cdots+l$
Write the sum forwards; $l$ is the last term.
$S_n=l+(l-d)+(l-2d)+\cdots+a$
The same sum backwards.
$2S_n=(a+l)+(a+l)+\cdots+(a+l)$
Add the two lines term by term: the $d$ 's cancel in every column.
$2S_n=n(a+l)\Rightarrow S_n=\tfrac n2(a+l)$
There are $n$ columns.
$S_n=\tfrac n2[2a+(n-1)d]$
Substitute $l=a+(n-1)d$ .

Worked exampleSum of the odd numbers between 100 and 200

Find $101+103+\cdots+199$ .

Goal: The total.
Strategy: Identify $a$ , $d$ , $l$ ; find $n$ ; use $\frac n2(a+l)$ .

$a=101,\ d=2,\ l=199$
Read off.
$101+(n-1)2=199\Rightarrow n=50$
$n$ th term $=l$ .
$S_{50}=\tfrac{50}2(101+199)=25\times300=7500$
Sum formula.

Answer

7500

✓ CheckAverage term

150

, times

50

terms

=7500

✓.

Worked exampleHow many terms to exceed 1000?

For $3+7+11+\cdots$ , find the least $n$ with $S_n>1000$ .

Goal: Smallest whole $n$ .
Strategy: Write $S_n$ , form a quadratic inequality, solve, round up.

$\tfrac n2[6+4(n-1)]=n(2n+1)$
$a=3$ , $d=4$ .
$2n^2+n-1000>0$
Inequality.
$n>\frac{-1+\sqrt{1+8000}}4=22.1$
Positive root of the quadratic.

Answer

n=23

✓ Check

S_{22}=22\times45=990

S_{23}=23\times47=1081

✓.

Common misconception $u_n$ versus $S_n$

u_n

is one term;

S_n

is the running total. Read carefully which is asked for.

Pause and think

Find $1+2+\cdots+100$ with the formula.

Summary

$u_n=a+(n-1)d$ .
$S_n=\frac n2(a+l)=\frac n2[2a+(n-1)d]$ (Gauss's pairing).

Common difference: The constant added each step.
Series: The sum of terms of a sequence.

HistoryThe Gauss story. A famous (and probably exaggerated) story says that the young Carl Friedrich Gauss, later one of the greatest mathematicians ever, was told by his teacher to add the numbers from

1

100

to keep him busy. He answered almost at once,

5050

, using the pairing trick above. That pairing is the proof of the sum formula.

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