Arithmetic Sequences and Series

Finding terms, sums, and identifying arithmetic patterns

SPEC 4.4 • SEQUENCES AND SERIES

Arithmetic Sequences

Definition

An arithmetic sequence is a sequence where each term is obtained by adding a constant value to the previous term. This constant is called the common difference, denoted by

d

.

Common Difference

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

nth Term Formula

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

Key

a

= first term,

d

= common difference,

n

= term number

Examples: 2, 7, 12, 17, ... has

a = 2

,

d = 5

30, 26, 22, 18, ... has

a = 30

,

d = - 4

Finding the nth Term Formula

Worked Example01

An arithmetic sequence begins 4, 9, 14, 19, ...

Find a formula for the nth term,

u_{n}

.

Step 1 Identify

a

and

d

First term:

a = 4

Common difference:

d = 9 - 4 = 5

Step 2 Substitute into formula

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

u_{n} = 4 + (n - 1) (5) = 4 + 5 n - 5 = 5n - 1

Finding a Specific Term

Worked Example02

An arithmetic sequence has first term 7 and common difference 3.

Find the 25th term.

a = 7

,

d = 3

,

n = 25

Using

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

:

u_{25} = 7 + (25 - 1) (3) = 7 + 24 \times 3 = 7 + 72 = 79

Practice

1 An arithmetic sequence has 3rd term equal to 14 and 8th term equal to 34.

(a) Find the common difference

d

(b) Find the first term

a

(c) Find the 50th term

Part a

u_{3} = a + 2 d = 14

and

u_{8} = a + 7 d = 34

Subtracting:

5 d = 20

, so

d = 4

Part b Using

a + 2 (4) = 14

:

a = 6

Part c

u_{50} = 6 + (50 - 1) (4) = 6 + 196 = 202

Arithmetic Series

Definition

An arithmetic series is the sum of the terms of an arithmetic sequence. We use

S_{n}

to denote the sum of the first

n

terms.

Sequence

3, 7, 11, 15, 19

The individual terms

Series

3 + 7 + 11 + 15 + 19 = 55

The sum of the terms

Notation

S_{n} = u_{1} + u_{2} + u_{3} + \dots + u_{n}

Deriving the Sum Formula

Consider the sum of an arithmetic series with first term

a

, common difference

d

, and

n

terms.

The Pairing Method

1 Write forwards:

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

2 Write backwards:

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

3 Add the two equations:

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

(n pairs)

4 Each pair sums to

[2 a + (n - 1) d]

, and there are

n

pairs:

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

, so

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

Alternative Form

If we let

l = a + (n - 1) d

be the last term, then

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

Sum Formulae

Version 1 (Formula Booklet)

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

Version 2 (using last term)

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

When to Use Which?

Version 1: When you know

a

,

d

, and

n

Version 2: When you know

a

,

l

(last term), and

n

Finding the Sum

Worked Example03

Find the sum of the first 20 terms of the arithmetic series: 3 + 7 + 11 + 15 + ...

a = 3

,

d = 4

,

n = 20

Using

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

:

S_{20} = \frac{20}{2} [2 (3) + (20 - 1) (4)]

S_{20} = 10 [6 + 76] = 10 \times 82 = 820

Practice

2 An arithmetic series has first term 5 and common difference 3.

(a) Find the sum of the first 25 terms.

(b) How many terms are needed for the sum to exceed 1000?

Part a

S_{25} = \frac{25}{2} [2 (5) + 24 (3)] = \frac{25}{2} [10 + 72] = \frac{25}{2} \times 82 = 1025

Part b Need

S_{n} > 1000

Test:

S_{24} = \frac{24}{2} [10 + 23 (3)] = 12 \times 79 = 948

(not enough)

S_{25} = 1025

(exceeds 1000)

Need 25 terms

Identifying Arithmetic Series

Key Skill

Recognise when a sum in sigma notation represents an arithmetic series.

How to Identify

A sigma notation represents an arithmetic series if the general term is linear in r (i.e., of the form

a r + b

).

Sigma NotationArithmetic?Reason

r = 1 \sum n (3 r + 2)

Yes ✓Linear in

r

r = 1 \sum n r^{2}

No ✗Quadratic in

r

r = 1 \sum n 2^{r}

No ✗Exponential in

r

Evaluating Using Arithmetic Properties

Worked Example04

(a) Show that

r = 1 \sum n (4 r - 1)

is an arithmetic series.

(b) Find the first term and common difference.

(c) Find the value when

n = 15

.

Part a The general term

4 r - 1

is linear in

r

, so it is arithmetic. ✓

Part b

u_{1} = 4 (1) - 1 = 3

, so

a = 3

. The coefficient of

r

gives

d = 4

.

Part c

S_{15} = \frac{15}{2} [2 (3) + 14 (4)] = \frac{15}{2} [6 + 56] = \frac{15}{2} \times 62 = 465

Real-World Application

Worked Example05

A theatre has 20 rows of seats. The first row has 15 seats, and each subsequent row has 2 more seats than the row in front.

(a) How many seats are in the 20th row?

(b) What is the total seating capacity?

a = 15

,

d = 2

,

n = 20

Part a

u_{20} = 15 + 19 (2) = 15 + 38 = 53

seats

Part b

S_{20} = \frac{20}{2} [2 (15) + 19 (2)] = 10 [30 + 38] = 10 \times 68 = 680

seats

Practice

3 Evaluate

r = 1 \sum 20 (5 r - 2)

Step 1 Identify this is arithmetic (linear in

r

)

First term:

a = 5 (1) - 2 = 3

Common difference:

d = 5

(coefficient of

r

)

Step 2 Use sum formula with

n = 20

S_{20} = \frac{20}{2} [2 (3) + 19 (5)] = 10 [6 + 95] = 10 \times 101 = 1010

Summary

Key Formulae

ConceptFormula

Common difference

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

nth term

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

Sum (version 1)

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

Sum (version 2)

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

Arithmetic in sigmaGeneral term is linear in

r

Key Exam Skills

Find common difference, nth term, and specific terms

Use sum formulae appropriately

Identify arithmetic series from sigma notation

Apply to real-world problems