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


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


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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