Difference between revisions of "Arithmetic series"

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is an arithmetic series whose value is 50.
 
is an arithmetic series whose value is 50.
 
==Formula==
 
==Formula==
To find the sum of an arithmetic sequence, we can write it out in two as so (<math>S</math> is the sum, <math>a</math> is the first term, <math>z</math> is the number of terms, and <math>d</math> is the common difference):
+
To find the sum of an arithmetic sequence, we can write it out in two as so (<math>S</math> is the sum, <math>a</math> is the first term, <math>z</math> is the last term, and <math>d</math> is the common difference):
 
<cmath>
 
<cmath>
 
S =  a  + (a+d) + (a+2d) + \ldots  + (z-d)  + z
 
S =  a  + (a+d) + (a+2d) + \ldots  + (z-d)  + z
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<cmath>2S = (a+z) + (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>, where <math>n</math> is the number of terms.
  
 
== Problems ==
 
== Problems ==

Revision as of 17:50, 17 February 2016

An arithmetic series is a sum of consecutive terms in an arithmetic sequence. For instance,

$2 + 6 + 10 + 14 + 18$

is an arithmetic series whose value is 50.

Formula

To find the sum of an arithmetic sequence, we can write it out in two as so ($S$ is the sum, $a$ is the first term, $z$ is the last term, and $d$ is the common difference): \[S =  a  + (a+d) + (a+2d) + \ldots  + (z-d)  + z\] Flipping the right side of the equation we get \[S = z   + (z-d) + (z-2d) + \ldots  +  (a+d)    + a\]

Now, adding the above two equations vertically, we get

\[2S = (a+z) + (a+z) + (a+z) + ... +(a+z) + (a+z)\]

This equals $2S = n(a+z)$, so the sum is $\frac{n(a+z)}{2}$, where $n$ is the number of terms.

Problems

Introductory Problems

Intermediate Problems

Olympiad Problem

See also

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