Arithmetic Sequence
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Logarithmic Spirals 
A sequence is an ordered list of numbers that often follow a particular pattern. For arithmetic sequences, the pattern is that numbers increase by a constant value. The first number in a sequence is usually denoted by a letter with a subscript of zero, the second with a subscript of 1, and so on.
A few example sequences are
 (1) 5,10,15,20,25...
 (2) 5,2,9,16,23,30
We can see that that the two sequences (1) and (2) are changing by 5 and 7, respectively.We examine sequence (1) to find this formula.
We begin by identifying the first term, 5, as and the following terms as .
We observe
From here we can reason that .
We can generalize further that the constant of increase does not need to be 5 but actually can be any value d.
Thus we have the expression
This means that the expression for sequence (2) is since the the initial value is 5 and the numbers change by 7.
Other Properties
Another property of arithmetic sequences is that any term is the average of its two neighbors. For example:
Looking at sequence (1),
In fact, the average of any two terms an even number of spaces apart is the value of the term in between.
Or more simply, the nth term of a sequence is the average of the (nt)th term and the (n+t)th term.
Summing the Sequence
The sum of an arithmetic sequence, which is called an arithmetic series, can be found with a general formula if the sequence terminates which means that it has a final value.
First, let's just look at some examples..
A simple sequence is . The series associated with this sequence is . This sum is simply .
Another sequence is The sum of these numbers is
Now let's find a general formula for the sum of a terminating arithmetic sequence. We first express the sequence in our already familiar way, denoting the sum of n terms as
(1):
Since and
we have that
And so we can rewrite the sequence as
(2):
By adding equations 1 and 2, we get
We can see that the first term contains , and the second term contains , so these two pieces cancel each other out. Similarly the third contains and the forth term contains . This goes on and therefore this rearrangement allows for mass cancellation of all terms of the form .
Now we only have the 's and the 's. Because there were n terms in both equations 1 and 2 that each contained either or , there are n 's and n 's.
So, we have .
Dividing by 2 gives us our desired equation for the sum:
Interactive Demonstration