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| − | An '''arithmetic series''' is a sum of consecutive terms in an [[arithmetic sequence]]. For instance,
| + | #REDIRECT[[Arithmetic sequence]] |
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| − | <math> 2 + 6 + 10 + 14 + 18 </math>
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| − | is an arithmetic series whose value is 50.
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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):
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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>
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| − | Now, adding the above two equations vertically, we get
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| − | <cmath>2S = (a+z) + (a+z) + (a+z) + ... + (a+z)</cmath>
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| − | This equals <math>2S = n(a+z)</math>, so the sum is <math>\frac{n(a+z)}{2}</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]]
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| − | === Intermediate Problems ===
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| − | *[[2003 AIME I Problems/Problem 2|2003 AIME I, Problem 2]]
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| − | === Olympiad Problem ===
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| − | == See also ==
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| − | * [[Series]]
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| − | * [[Summation]]
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| − | {{stub}}
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