Arithmetic Sequences and Sums
Sequence
A Sequence is a set of things (usually numbers) that are in order.
Arithmetic Sequence
In an Arithmetic Sequence the difference between one term and the next is a constant.
In other words, you just add the same value each time ... on to infinity.
Example:
| 1, 4, 7, 10, 13, 16, 19, 22, 25, ... |
This sequence has a difference of 3 between each number.
In General you 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")
And you can make the rule by:
xn = a + d(n-1)
(We use "n-1" because d is not used in the 1st term).
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:
xn = a + d(n-1)
= 3 + 5(n-1)
= 3 + 5n - 5
= 5n - 2
So, the 4th term is:
x4 = 5·4 - 2 = 18
Is that right? Check for yourself!
Arithmetic Sequences are sometimes called Arithmetic Progressions (A.P.’s)
Summing an Arithmetic Series
To sum up the terms of an arithmetic sequence like this:
a + (a+d) + (a+2d) + (a+3d) + ...
you can use this formula:
What is that funny symbol? It is called Sigma Notation
 |
This symbol (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:
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
Check: why don't you add up the terms yourself, and see if it comes to 145
Why Does the Formula Work?
I want to show you 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 the terms "underneath" each other together (i.e. add the 1st terms together, the 2nd terms together, etc):
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:
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