1.04PureCore

Arithmetic Sequences

Q: What is the difference between a sequence and a series?

A sequence is a list of terms; a series is the sum of the terms of a sequence.

Q: Which sum formula should I use?

Use n2(2a + (n-1)d) when you know a and d, and n2(a + l) when you know the first and last terms.

Q: How do I show a sequence is arithmetic?

Show that u_n+1 - u_n is the same constant for every n, not just for the first few terms.

Q: Can the common difference be negative or a fraction?

Yes. A negative d gives a decreasing sequence, and any real number is allowed.

Seats in a theatre that increase by three per row, a savings plan that adds £40 a month, a stack of cans with one fewer in each layer — each is an $\textbf{arithmetic sequence}$ , where every term differs from the last by the same amount. Two formulas, one for the $n$ th term and one for the sum, answer almost every question.

40 min Sequences and Series

The big picture

Arithmetic sequences are the simplest sequences with structure, which makes them the right place to learn the language of the whole topic: $u_{n}$ for terms, recurrence relations like $u_{n+1} = u_{n} + d$ , and sigma notation for sums. The sum formula has a lovely origin — pair the first term with the last, the second with the second-last, and every pair has the same total — and understanding that pairing argument is more useful than memorising the result, because it is exactly the kind of reasoning OCR asks you to reproduce as a proof. Geometric sequences, in the next lesson, use the same notation with multiplication instead of addition.

What you'll be able to do

Describe sequences using nth-term formulas and recurrence relations
Use sigma notation for sums
Find the $n$ th term of an arithmetic sequence, $u_{n} = a + (n-1)d$
Prove and use $S_{n} = \tfrac{n}{2}\big(2a + (n-1)d\big)$

Describing sequences

A sequence can be defined by a $\textbf{position-to-term}$ rule, such as $u_{n} = 3n + 2$ , which gives any term directly: $u_{10} = 32$ .

Or by a $\textbf{recurrence relation}$ , which gives each term from the one before, together with a starting value: $u_{1} = 5$ , $u_{n+1} = 2u_{n} - 1$ generates $5, 9, 17, 33, \dots$

A sequence is $\textbf{increasing}$ if $u_{n+1} > u_{n}$ for all $n$ , $\textbf{decreasing}$ if $u_{n+1} < u_{n}$ for all $n$ , and $\textbf{periodic}$ if its terms repeat in a cycle — for example $u_{n+1} = \dfrac{1}{u_{n}}$ with $u_{1} = 2$ gives $2, \tfrac12, 2, \tfrac12, \dots$ with period 2.

\sum_{r=1}^{n} u_{r} = u_{1} + u_{2} + \dots + u_{n}

Sigma notation.

Tip — Count the terms in a sigma sum carefully: $\sum_{r=3}^{10}$ has $10 - 3 + 1 = 8$ terms, not 7.

The nth term

An $\textbf{arithmetic sequence}$ has first term $a$ and $\textbf{common difference}$ $d$ , so each term is $d$ more than the previous one: $u_{n+1} = u_{n} + d$ .

Getting from the first term to the $n$ th takes $n - 1$ steps, each adding $d$ , so $u_{n} = a + (n-1)d$ .

Arithmetic sequences grow (or shrink) $\textit{linearly}$ : plotting $u_{n}$ against $n$ gives points on a straight line of gradient $d$ .

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

nth term of an arithmetic sequence.

Example 1The 4th term of an arithmetic sequence is 23 and the 11th term is 58. Find

a

d

and the first term greater than 200.

a + 3d = 23

and

a + 10d = 58

2Subtracting:

7d = 35

, so

d = 5

, and

a = 8

3Need

8 + 5(n-1) > 200

, so

5(n-1) > 192

and

n - 1 > 38.4

4The smallest integer is

n = 40

, giving

u_{40} = 8 + 5 \times 39

Answer

a = 8

d = 5

;

u_{40} = 203

is the first term above 200.

Summing an arithmetic series

Write the sum forwards and backwards: $S_{n} = a + (a + d) + \dots + \big(a + (n-1)d\big)$ and $S_{n} = \big(a + (n-1)d\big) + \dots + (a + d) + a$ .

Adding the two lines term by term, every column totals $2a + (n-1)d$ , and there are $n$ columns. So $2S_{n} = n\big(2a + (n-1)d\big)$ .

If you know the last term $l = a + (n-1)d$ , the same result is $S_{n} = \tfrac{n}{2}(a + l)$ : the number of terms times the average of the first and last.

S_{n} = \dfrac{n}{2}\big(2a + (n-1)d\big)

Sum of the first n terms.

S_{n} = \dfrac{n}{2}(a + l)

Using the last term.

Example 1Find the sum of all the multiples of 7 between 100 and 500.

1First multiple above 100:

105

. Last below 500:

497

2Number of terms:

\dfrac{497 - 105}{7} + 1 = 57

S = \dfrac{57}{2}(105 + 497) = \dfrac{57}{2} \times 602

Answer

17\,157

Why this works

The pairing proof is the one Gauss is said to have spotted as a schoolboy to add 1 to 100 in seconds: fifty pairs each summing to 101. OCR may ask you to reproduce this proof, so learn the argument, not just the formula.

Problems in context

In applied questions, decide first what $a$ , $d$ and $n$ represent, and whether the question wants a term (one value) or a sum (a running total).

When solving $S_{n} = k$ for $n$ , you get a quadratic. The number of terms must be a positive integer, so reject negative or non-integer roots — or interpret them, if the question asks "when does the total first exceed...".

Example 1A runner trains 3 km on day 1 and increases the distance by 0.5 km each day. On which day does her total distance first exceed 100 km?

a = 3

d = 0.5

S_{n} = \dfrac{n}{2}\big(6 + 0.5(n-1)\big) = \dfrac{n}{4}(n + 11)

2Solve

\dfrac{n}{4}(n + 11) = 100

n^{2} + 11n - 400 = 0

n = \dfrac{-11 + \sqrt{121 + 1600}}{2} = \dfrac{-11 + 41.48}{2} = 15.24

4Check:

S_{15} = \tfrac{15}{4}(26) = 97.5

and

S_{16} = \tfrac{16}{4}(27) = 108

AnswerDay 16

Tip — Always check the integer values either side of a non-integer root. It confirms the answer and earns the final mark.

Think like an examiner

Set up simultaneous equations from two given terms rather than guessing $a$ and $d$ .
Be ready to prove the sum formula using the forwards-and-backwards pairing.
Count terms in a sigma sum or a range of multiples precisely.
Justify the integer answer when a quadratic in $n$ has a non-integer root.

Common misconceptions

Using

u_{n} = a + nd

There are

n - 1

steps from term 1 to term

n

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

Confusing

u_{n}

with

S_{n}

in context questions.

A single day, row or payment is a term; a running total is a sum.

Counting

\sum_{r=5}^{20}

as 15 terms.

It has

20 - 5 + 1 = 16

terms.

Rounding

n = 15.24

down to 15 for "first exceeds".

S_{15}

is still below the target, so the answer is

n = 16

Arithmetic sequences

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

nth term.

S_{n} = \tfrac{n}{2}\big(2a + (n-1)d\big)

Sum of n terms.

S_{n} = \tfrac{n}{2}(a + l)

Sum using last term.

\sum_{r=1}^{n} r = \tfrac12 n(n+1)

Sum of the first n integers.

Stretch yourself

The sum of the first $n$ terms of a sequence is given by $S_{n} = 3n^{2} + 2n$ for all $n \ge 1$ . Show that the sequence is arithmetic and find its first term and common difference.

Hint — Each term is the difference of consecutive sums: $u_{n} = S_{n} - S_{n-1}$ for $n \ge 2$ , and $u_{1} = S_{1}$ .

Reveal solution

SolutionFor

n \ge 2

u_{n} = S_{n} - S_{n-1} = 3n^{2} + 2n - \big(3(n-1)^{2} + 2(n-1)\big) = 3n^{2} + 2n - (3n^{2} - 6n + 3 + 2n - 2) = 6n - 1

. Also

u_{1} = S_{1} = 5

, and

6(1) - 1 = 5

, so

u_{n} = 6n - 1

for every

n \ge 1

. Then

u_{n+1} - u_{n} = 6

, a constant, so the sequence is arithmetic with

a = 5

and

d = 6

. Check with the sum formula:

\tfrac{n}{2}\big(10 + 6(n-1)\big) = \tfrac{n}{2}(6n + 4) = 3n^{2} + 2n

✓.

\blacksquare

Questions students ask

What is the difference between a sequence and a series?

A sequence is a list of terms; a series is the sum of the terms of a sequence.

Which sum formula should I use?

Use

\tfrac{n}{2}(2a + (n-1)d)

when you know

a

and

d

, and

\tfrac{n}{2}(a + l)

when you know the first and last terms.

How do I show a sequence is arithmetic?

Show that

u_{n+1} - u_{n}

is the same constant for every

n

, not just for the first few terms.

Can the common difference be negative or a fraction?

Yes. A negative

d

gives a decreasing sequence, and any real number is allowed.

Key takeaways

Sequences can be defined by an nth-term rule or a recurrence relation.
Arithmetic: constant difference $d$ ; $u_{n} = a + (n-1)d$ .
$S_{n} = \tfrac{n}{2}(2a + (n-1)d) = \tfrac{n}{2}(a + l)$ , proved by pairing terms.
Sigma notation $\sum_{r=1}^{n}$ compactly writes a sum.
In context, decide whether a term or a sum is needed, and check integer answers.