Difference between revisions of "Arithmetic series"

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An '''arithmetic series''' is a sum of consecutive terms in an [[arithmetic sequence]].  For instance,
 
An '''arithmetic series''' is a sum of consecutive terms in an [[arithmetic sequence]].  For instance,
  
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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):
S =  a + (a+d) + (a+2d) + ... + (a+(n-1)d)
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<div style="text-align:center;"><math>\displaystyle S =  a + (a+d) + (a+2d) + ... + (a+(n-1)d)</math>
  
S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a
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<math>\displaystyle S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a</math></div>
  
 
Now, adding vertically and shifted over one, we get
 
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)
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<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
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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>.
  
2S = n(2a+(n-1)d), so the sum is <math>\displaystyle \frac{n}{2} (2a+(n-1)d</math>
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== Problems ==
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=== Introductory Problems ===
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* [[2006_AMC_10A_Problems/Problem_9 | 2006 AMC 10A, Problem 9]]
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*[[2006 AMC 12A Problems/Problem 12 | 2006 AMC 12A, Problem 12]]
  
== Example Problems ==
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=== Intermediate Problems ===
=== Introductory Problems ===
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*[[2003 AIME I Problems/Problem 2|2003 AIME I, Problem 2]]
* [[2006_AMC_10A_Problems/Problem_9 | 2006 AMC 10A Problem 9]]
 
  
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=== Olympiad Problem ===
  
 
== See also ==
 
== See also ==
 
* [[Series]]
 
* [[Series]]
 
* [[Summation]]
 
* [[Summation]]
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Revision as of 15:55, 12 March 2007

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 as so (S is the sum, a is the first term, n is the number of terms, and d is the common difference):

$\displaystyle S =  a + (a+d) + (a+2d) + ... + (a+(n-1)d)$ $\displaystyle S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a$

Now, adding vertically and shifted over one, we get

$\displaystyle 2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)$

This equals $\displaystyle 2S = n(2a+(n-1)d)$, so the sum is $\displaystyle \frac{n}{2} (2a+(n-1)d$.

Problems

Introductory Problems

Intermediate Problems

Olympiad Problem

See also

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