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 $a_0\,$ and the following terms $10, 15, 20... \,$ as $a_1, a_2, a_3,...\,$.

We observe

$a_1 = a_0 + 5\,$
$a_2 = a_1 + 5 = a_0 + 5*2\,$
$a_3 = a_2 + 5 = a_0 + 5*3\,$
$a_4 = a_3 + 5 = a_0 + 5*4\,$

From here we can reason that $a_n = a_0 + 5 * n\,$.

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

$a_n=a_0+d*n\,$

This means that the expression for sequence (2) is $b_n=5+ (-7)n$ 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),

$10 = \frac{5+15}{2} = 10\,$

In fact, the average of any two terms an even number of spaces apart is the value of the term in between.

$a_n = a_0 + d * n = \frac{(a_0 + d * {(n-t)})+(a_0 +d * {(n+t)})}{2} = \frac{a_{(n-t)}+a_{(n+t)}}{2}$

Or more simply, the nth term of a sequence is the average of the (n-t)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 $1,2,3,4,5,6,7$. The series associated with this sequence is $1+2+3+4+5+6+7$. This sum is simply $28$.

Another sequence is $2,4,6,8,10,12$ The sum of these numbers is $2+4+6+8+10+12=42$

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 $S_n = a_0 +a_1+... a_{n-1}.$

(1):$S_n=a_0+(a_0+d)+(a_0+2d)+\cdots+(a_0+(n-2)d)+(a_0+(n-1)d)$

Since $a_{n-1}=a_0+(n-1)d$ and $a_{n-1}=a_{n-2}+d=a_0+(n-2)d$

we have that

$a_0=(a_{n-1}-(n-1)d), a_1=(a_{n-1}-(n-2)d), \cdots$

And so we can rewrite the sequence as

(2):$S_n=(a_{n-1}-(n-1)d)+(a_{n-1}-(n-2)d)+\cdots+(a_{n-1}-2d)+(a_{n-1}-d)+a_{n-1}.$

By adding equations 1 and 2, we get

$2S_n=((a_{n-1}-(n-1)d)+(a_0+(n-1)d))+((a_{n-1}-(n-2)d)+(a_0+(n-2)d))+\cdots+(a_{n-1}-2d)+(a_{n-1}-d)+(a_{n-1}+a_0).$

We can see that the first term contains $-(n-1)d$, and the second term contains $(n-1)d$, so these two pieces cancel each other out. Similarly the third contains $-(n-2)d$ and the forth term contains $(n-2)d$. This goes on and therefore this rearrangement allows for mass cancellation of all terms of the form $(n-m)d$.

Now we only have the $a_{n-1}$'s and the $a_0$'s. Because there were n terms in both equations 1 and 2 that each contained either $a_{n-1}$ or $a_0$, there are n $a_{n-1}$'s and n $a_0$'s.

So, we have $2S_n= n*a_{n-1}+n*a_0$.

Dividing by 2 gives us our desired equation for the sum:

$S_n=\frac{n( a_0 + a_{n-1})}{2}$

Interactive Demonstration

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