Difference between revisions of "Arithmetic series"
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To find the sum of an arithmetic sequence, we can write it out as so (S is the sum, a is the first term, n is the number of terms, and d is the common difference): | To find the sum of an arithmetic sequence, we can write it out as so (S is the sum, a is the first term, n is the number of terms, and d is the common difference): | ||
| − | < | + | <cmath>\begin{align*} |
| − | + | S &= a + (a+d) + (a+2d) + ... + (a+(n-1)d) \\ | |
| − | + | S &= (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a | |
| + | \end{align*}</cmath> | ||
Now, adding vertically and shifted over one, we get | Now, adding vertically and shifted over one, we get | ||
| − | < | + | <cmath>2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)</cmath> |
| − | This equals <math> | + | This equals <math>2S = n(2a+(n-1)d)</math>, so the sum is <math>\frac{n}{2} (2a+(n-1)d)</math>. |
== Problems == | == Problems == | ||
Revision as of 13:31, 24 April 2009
An arithmetic series is a sum of consecutive terms in an arithmetic sequence. For instance,
is an arithmetic series whose value is 50.
To find the sum of an arithmetic sequence, we can write it out as so (S is the sum, a is the first term, n is the number of terms, and d is the common difference):
Now, adding vertically and shifted over one, we get
![\[2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)\]](http://latex.artofproblemsolving.com/b/f/e/bfe3c605cd87a957d64c1777860d0333b4fa885e.png)
This equals 
, so the sum is 
.
Contents
Problems
Introductory Problems
Intermediate Problems
Olympiad Problem
See also
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