Arithmetic series: Difference between revisions
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Now, adding the above two equations vertically, we get | Now, adding the above two equations vertically, we get | ||
<cmath>2S = (a+z) + (a+z) + (a+z) + ... + (a+z)</cmath> | <cmath>2S = (a+z) + (a+z) + (a+z) + ... +(a+z) + (a+z)</cmath> | ||
This equals <math>2S = n(a+z)</math>, so the sum is <math>\frac{n(a+z)}{2}</math>. | This equals <math>2S = n(a+z)</math>, so the sum is <math>\frac{n(a+z)}{2}</math>. | ||
Revision as of 19:31, 19 September 2015
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 in two as so (
is the sum, 
is the first term, 
is the number of terms, and 
is the common difference):
![\[S = a + (a+d) + (a+2d) + \ldots + (z-d) + z\]](http://latex.artofproblemsolving.com/7/f/8/7f8721e94b3efdffa088da3d04c8b22844eda638.png)
Flipping the right side of the equation we get
![\[S = z + (z-d) + (z-2d) + \ldots + (a+d) + a\]](http://latex.artofproblemsolving.com/5/9/d/59da689c1428ebaefe9e745a75c08325b348eb22.png)
Now, adding the above two equations vertically, we get
![\[2S = (a+z) + (a+z) + (a+z) + ... +(a+z) + (a+z)\]](http://latex.artofproblemsolving.com/0/5/4/0543bf3046b2d2ff0dad861f0b23cc8b20662057.png)
This equals 
, so the sum is 
.
Problems
Introductory Problems
Intermediate Problems
Olympiad Problem
See also
This article is a stub. Help us out by expanding it.
