Difference between revisions of "Arithmetic series"

(Changes to formula)
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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 number of terms, and <math>d</math> is the common difference):
 
<cmath>
 
<cmath>
S =  a  + (a+d) + (a+2d) + ... + (z-d)  + z
+
S =  a  + (a+d) + (a+2d) + \ldots + (z-d)  + z
 
</cmath>
 
</cmath>
 
Flipping the right side of the equation we get
 
Flipping the right side of the equation we get
 
<cmath>
 
<cmath>
S = z  + (z-d) + (z-2d) +...  +  (a+d)    + a
+
S = z  + (z-d) + (z-2d) + \ldots  +  (a+d)    + a
 
</cmath>
 
</cmath>
  

Revision as of 20:30, 19 September 2015

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.

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 number of terms, 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

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

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

Problems

Introductory Problems

Intermediate Problems

Olympiad Problem

See also

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