D1PureCore

Arithmetic Sequences and Series

Q: Why is it (n-1)d and not nd in the n-th term?

Because you start at the first term a and only add d to move between terms. Reaching the n-th term takes n-1 of those steps, so you add d exactly n-1 times.

Q: Which sum formula should I use?

Use S_n = n2(a + L) when you know the last term L. Use S_n = n2[2a + (n-1)d] when you only have the first term, common difference and number of terms.

An arithmetic sequence goes up (or down) in equal steps. Two formulas do everything: one for any term, and one — discovered by a nine-year-old — for adding up a whole run of them at once.

28 min Video by TLMaths Sequences and Series

The big picture

Arithmetic sequences are the simplest kind of pattern: add the same amount each time. That simplicity makes them the perfect place to meet two ideas that run through all of series work — a formula for the $n$ -th term (so you never list out a sequence to find its 100th value) and a formula for a $series$ (the sum of many terms). The sum formula has a famous origin: the young Gauss reportedly added $1 + 2 + \dots + 100$ in seconds by pairing the ends. That pairing trick is not a party piece — it is the actual derivation, and understanding it means you never have to memorise the formula blindly.

A-Level Maths D4-01: Introducing Arithmetic SequencesWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Identify an arithmetic sequence and its common difference
Use the $n$ -th term formula $a + (n - 1) d$
Derive and use the sum formula for an arithmetic series
Solve problems where a term or sum is given and $n$ , $a$ or $d$ is unknown

Sequences with a constant step

An $arithmetic sequence$ increases by the same fixed amount each term — the $common difference$ $d$ . The first term is $a$ . So $3, 7, 11, 15, \dots$ has $a = 3$ and $d = 4$ .

To reach the $n$ -th term you start at $a$ and add $d$ a total of $(n - 1)$ times — not $n$ times, because the first term needs no steps. That single observation gives the $n$ -th term formula.

u_{n} = a + (n - 1) d

Start at

a

, then add

d

exactly

(n - 1)

times.

1Here

a = 3

d = 4

u_{20} = 3 + (20 - 1) (4) = 3 + 76 = 79

Answer

u_{20} = 79

Tip — The $(n - 1)$ trips up everyone at least once: the 1st term uses zero steps, the 10th uses nine. Count the gaps, not the terms.

Summing a series — the pairing trick

A $series$ is the sum of the terms of a sequence. Gauss’s insight: write the sum forwards and again backwards, and add the two lines. Every vertical pair adds to the same total ( $a + L$ , first plus last), and there are $n$ pairs — so twice the sum is $n (a + L)$ .

That gives $S_{n} = \frac{n}{2} (a + L)$ , where $L$ is the last term. Substituting $L = a + (n - 1) d$ gives the fully expanded version.

S_{n} = \frac{n}{2} (a + L)

Half of (number of terms) × (first + last).

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

The same, with

L

written out.

a = 3

d = 4

n = 20

2Use

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

S_{20} = \frac{20}{2} [6 + 19 (4)]

= 10 [6 + 76] = 10 \times 82 = 820

Answer

S_{20} = 820

Why this works

The two sum formulas are the same fact. Use $\frac{n}{2} (a + L)$ when you already know the last term; use $\frac{n}{2} [2 a + (n - 1) d]$ when you only know $a$ , $d$ and $n$ .

Working backwards to a hidden unknown

Many questions give you a term or a sum and ask for $n$ , $a$ or $d$ . The strategy is always the same: substitute what you know into the right formula and solve — sometimes forming a pair of simultaneous equations, sometimes a quadratic in $n$ .

1Use

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

with

a = 5

d = - 2

S_{n} = 0

0 = \frac{n}{2} [10 - 2 (n - 1)] = \frac{n}{2} [12 - 2 n]

3So

n = 0

(rejected) or

12 - 2 n = 0 \Rightarrow n = 6

Answer

n = 6

Think like an examiner

The $n$ -th term uses $(n - 1) d$ , not $n d$ — the single most common arithmetic-series error.
Choose the sum formula that matches what you are given: $\frac{n}{2} (a + L)$ if the last term is known, otherwise the $2 a + (n - 1) d$ version.
When solving for $n$ , you often get a quadratic — reject any non-positive or non-integer value with a brief reason.
Two unknowns (say $a$ and $d$ ) usually mean two given facts → set up and solve simultaneous equations.

Common misconceptions

Using

u_{n} = a + n d

for the

n

-th term.

It is

a + (n - 1) d

: the first term takes zero steps of

d

Mixing up the sequence (a list) and the series (its sum).

A sequence lists terms; a series adds them. The sum formulas are for the series.

Applying

\frac{n}{2} (a + L)

without knowing the last term

L

L

is unknown, use

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

instead.

Accepting a negative or fractional

n

from a quadratic.

The number of terms must be a positive integer — reject other solutions.

Arithmetic formulas

u_{n} = a + (n - 1) d

n

-th term.

S_{n} = \frac{n}{2} (a + L)

Sum (last term known).

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

Sum (from

a

d

n

Stretch yourself

The sum of the first $n$ terms of an arithmetic series is $S_{n} = 2 n^{2} + 3 n$ . Find the first term and the common difference.

Hint — The first term is $S_{1}$ . The second term is $S_{2} - S_{1}$ . The common difference is the gap between consecutive terms.

Questions students ask

What is the difference between a sequence and a series?

A sequence is an ordered list of terms; a series is what you get when you add those terms together. The

n

-th term formula describes the sequence; the sum formulas describe the series.

Why is it

(n - 1) d

and not

n d

in the

n

-th term?

Because you start at the first term

a

and only add

d

to move between terms. Reaching the

n

-th term takes

n - 1

of those steps, so you add

d

exactly

n - 1

times.

Which sum formula should I use?

Use

S_{n} = \frac{n}{2} (a + L)

when you know the last term

L

. Use

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

when you only have the first term, common difference and number of terms.

How do I find

n

when I know the sum?

Substitute into the sum formula and solve. This usually produces a quadratic in

n

; take the positive integer solution and reject the rest.

Key takeaways

An arithmetic sequence adds a constant $d$ each term.
$n$ -th term: $a + (n - 1) d$ — count the $n - 1$ gaps, not the terms.
Sum: $\frac{n}{2} (a + L)$ or $\frac{n}{2} [2 a + (n - 1) d]$ (Gauss’s pairing).
A sequence is a list; a series is its sum.
Given a term or sum, substitute and solve — often a quadratic in $n$ .

How this fits the course

Build on

Sequences (GCSE)soon

Leads to

Geometric Sequences and Series

Test yourself

Ready to lock in Arithmetic Sequences and Series? Pick a mode and earn XP & Dobloons.