Topic 5: Arithmetice Sequence and Sums

Arithmetic Sequence

In an Arithmetic Sequence the difference between one term and the next is a constant.
In other words, we just add the same value each time infinitely.

For example;

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number.

In general we could write an arithmetic sequence like this:

{ a, a+d, a+2d, a+3d, ... }

Where:

• a is the first term, and
• d is the difference between the terms (called the "common difference")

For example; (Continued)

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

Has:

• a = 1 (First term), and
• d = 3 (The "common difference" between terms)

and we get:

{ a, a + d, a + 2d, a + 3d, ... }

{ 1, 1 + 3, 1 + 2 × 3, 1 + 3 × 3, ... }

{ 1, 4, 7, 10, ... }

Rule

We can write an Arithmetic Sequence as a rule:

x_n = a + d(n-1)

We use "n-1" because d is not used in the 1st term.

For example: Write the Rule, and calculate the 4th term for

3, 8, 13, 18, 23, 28, 33, 38, ...

This sequence has a difference of 5 between each number.

The values of a and d are:

• a = 3 (the first term)
• d = 5 (the "common difference")

The Rule can be calculated:

x_n = a + d(n-1)

3 + 5(n-1)

3 + 5n - 5

5n - 2

So, the 4th term is:

x₄ = 5×4 - 2 = 18

Arithmetic Sequences are sometimes called Arithmetic Progressions.


Summing an Arithmetic Series

To sum up the terms of this arithmetic sequence:

a + (a+d) + (a+2d) + (a+3d) + ...

use this formula:

What is that funny symbol? It is called Sigma Notation

∑ (called Sigma) means "sum up"

And below and above it are shown the starting and ending values:

It says "Sum up n where n goes from 1 to 4. Answer = 10"

Here is how to use it:

For example: Add up the first 10 terms of the arithmetic sequence

{ 1, 4, 7, 10, 13, ... }

The values of a, d and n are:

• a = 1 (the first term)
• d = 3 (the "common difference" between terms)
• n = 10 (how many terms to add up)

So:

Becomes:

= 5(2 + 9 × 3) = 5(29) = 145

Why Does the Formula Work?

Let's see why the formula works, because we get to use an interesting "trick" which is worth knowing.

First, we will call the whole sum "S":

S = a + (a + d) + ... + (a + (n-2)d) + (a + (n-1)d)

Next, rewrite S in reverse order:

S = (a + (n-1)d) + (a + (n-2)d) + ... + (a + d) + a

Now add those two, term by term:

S = a + (a+d) + ... + (a + (n-2)d) + (a + (n-1)d) 

S = (a + (n-1)d) + (a + (n-2)d) + ... + (a + d) + a

2S = (2a + (n-1)d) + (2a + (n-1)d) + ... + (2a + (n-1)d) + (2a + (n-1)d)

Each term is the same! And there are "n" of them so...

2S = n × (2a + (n-1)d)

Now, just divide by 2 and we get:

S = (n/2) × (2a + (n-1)d)

Which is our formula: