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.
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==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 last term, and <math>d</math> is the common difference):
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<cmath>
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S =  a  + (a+d) + (a+2d) + \ldots  + (z-d)  + z
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</cmath>
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Flipping the right side of the equation we get
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<cmath>
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S = z  + (z-d) + (z-2d) + \ldots  +  (a+d)    + a
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</cmath>
  
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):
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Now, adding the above two equations vertically, we get
<div style="text-align:center;"><math>\displaystyle S =  a + (a+d) + (a+2d) + ... + (a+(n-1)d)</math>
 
  
<math>\displaystyle S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a</math></div>
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<cmath>2S = (a+z) + (a+z) + (a+z) + ... +(a+z) + (a+z)</cmath>
  
Now, adding vertically and shifted over one, we get
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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.
 
 
<div style="text-align:center;"><math>\displaystyle 2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)</math></div>
 
 
 
This equals <math>\displaystyle 2S = n(2a+(n-1)d)</math>, so the sum is <math>\displaystyle \frac{n}{2} (2a+(n-1)d</math>.
 
  
 
== Problems ==
 
== Problems ==

Revision as of 18:50, 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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