Arithmetic Sequences
An arithmetic sequence is a sequence of numbers in which the difference between each term an and the preceding term an-1 is constant. $$ a_n = a_{n-1}+d $$ Here, n denotes a natural number.
This constant difference is called the common difference of the sequence.
$$ d=a_n-a_{n-1} $$
The value of d determines the behavior of the sequence:
- d>0
the sequence is increasing - d<0
the sequence is decreasing - d=0
the sequence is constant
An arithmetic sequence can have either a finite or an infinite number of terms. When the number of terms is finite, we refer to the first and last elements as the first term and the last term.
Why is it called an arithmetic sequence? The name comes from a key property: each term an is the arithmetic mean of its neighbors. $$ a_n = \frac{a_{n-1}+a_{n+1}}{2} $$
How do you compute a term?
Each term can be determined using the terms next to it:
- From the previous term
Add the common difference to the preceding term. \[ a_n = a_{n-1} + d \] - From the next term
Subtract the common difference from the following term. \[ a_n = a_{n+1} - d \]
In practice, once you know the first term and the common difference, the entire sequence is fixed.
Example
Consider the sequence
$$ 3 \ , \ 5 \ , \ 7 \ , \ 9 \ , \ 11 \ , \ 13 \ \dots $$
This is an arithmetic sequence with infinitely many terms.
The common difference is
$$ d = 2 $$
Note. Each term is obtained by adding 2 to the previous one $$ 5 = 3 +2 \\ 7 = 5+2 \\ 9 = 7+2 \\ 11 = 9 +2 \\ 13 = 11+2 \\ $$
Since d>0, the sequence is increasing.
Example 2
Consider the sequence
$$ 21 \ , \ 18 \ , \ 15 \ , \ 12 \ , \ 9 \ , \ 6 $$
This is a finite arithmetic sequence.
The first term is 21 and the last term is 6.
The common difference is
$$ d = -3 $$
Since d<0, the sequence is decreasing.
Example 3
Consider the sequence
$$ 5 \ , \ 5 \ , \ 5 \ , \ 5 \ , \ 5 \ , \ 5 \ , \ \dots $$
This is a constant arithmetic sequence.
In this case, the common difference is
$$ d=0 $$
Key properties
Here are some useful results that make arithmetic sequences especially easy to work with:
- The general term can be written explicitly as $$ a_n = a_1 + (n-1) \cdot d $$
Example. Let d=2 and a1=3 $$ 3 \ , \ 5 \ , \ 7 \ , \ 9 \ , \ 11 \ , \ 13 \ \dots $$ The fourth term is $$ a_4 = 3+(4-1) \cdot 2 = 9 $$ Why does this work? The difference between consecutive terms is constant $$ a_2 - a_{1} = d \\ a_3 - a_{2} = d \\ \vdots \\ a_n - a_{n-1} = d \\ $$ Adding all these equalities gives $$ a_n - a_1 = (n-1) \cdot d $$
- Any two terms are related by $$ a_x = a_y + (x-y) \cdot d $$
Example. In the sequence $$ 3 \ , \ 5 \ , \ 7 \ , \ 9 \ , \ 11 \ , \ 13 \ \dots $$ we have $$ a_2 = a_4 + (2-4) \cdot 2 $$ $$ 5 = 9 -4 = 5 $$
- In the first n terms, the sum of two terms equidistant from the ends is constant and equal to a1+an
Example. $$ 3 \ , \ 5 \ , \ 7 \ , \ 9 \ , \ 11 \ , \ 13 $$ The endpoints give $$ a_1 + a_6 = 3+13 = 16 $$ For example $$ a_2 + a_5 = 5+11 = 16 $$
- The sum of the first n terms is given by $$ S_n = n \cdot \frac{a_1+a_n}{2} $$
Example. $$ 3 \ , \ 5 \ , \ 7 \ , \ 9 \ , \ 11 \ , \ 13 $$ $$ 3+5+7+9+11+13 = 48 $$ Using the formula $$ S_6 = 6 \cdot \frac{3+13}{2} = 48 $$
And so on.
