Editing Arithmetic progression (section)

==Sum==

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{| style="background-color:white; width:220px;"
| 2 || + || 5 || + || 8 || + || 11 || + || 14 || = || 40
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| 14 || + || 11 || + || 8 || + || 5 || + || 2 || = || 40
|-
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| 16 || + || 16 || + ||  16 || + ||  16 || + ||  16 || = || 80
|}
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Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.
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A finite arithmetic series is a [[Summation|sum]] of the members of a finite arithmetic progression; it is commonly abbreviated to simply ''arithmetic series'' when the context is appropriate. It can be expressed as

:<math>a+(a+d)+(a+2d)+\cdots+(a+(n-1)d) = \sum_{k=1}^n a+(k-1)d. </math>

For example, consider the sum:

:<math>2 + 5 + 8 + 11 + 14 = 40 </math>

This sum can be found quickly by taking the number ''n'' of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:

:<math>\frac{n(a_1 + a_n)}{2}</math>

In the case above, this gives the equation:

:<math>2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.</math>

This formula works for any arithmetic progression of real numbers beginning with <math>a_1</math> and ending with <math>a_n</math>. For example,

:<math>\left(-\frac{3}{2}\right) + \left(-\frac{1}{2}\right) + \frac{1}{2} = \frac{3\left(-\frac{3}{2} + \frac{1}{2}\right)}{2} = -\frac{3}{2}.</math>

If the arithmetic series is an [[series (mathematics)|infinite series]], it [[divergent series|diverges]]<ref>{{cite book |author1=Fox, Huw|author2=Bolton, Bill|url=https://www.sciencedirect.com/science/article/abs/pii/B9780750655446500084 |title=Mathematics for Engineers and Technologists |publisher=Elsevier Ltd. |year=2002 |isbn=978-0-7506-5544-6 |url-access=limited}}</ref><ref>{{cite book |author1=Mcdougal, Holt |author2=Chard, David J. |author3=Hall, Earlene J. |author4=  Kennedy, Paul A. |author5=Leinwand, Steven J. |author6=Renfro, Freddie L. |author7=Seymour, Dale G. |author8=Waits, Bert K. |title=Holt Algebra 2 |publisher=Holt Rinehart & Winston |date=October 26, 2007 |isbn=978-0030358296}}</ref> (except when it is the trivial case of the sum of a series of zeroes). However, in some contexts, values can be assigned to series such as [[1 + 2 + 3 + 4 + ⋯]], using summation methods.

=== Derivation ===
[[File:Animated proof for the formula giving the sum of the first integers 1+2+...+n.gif|thumb|Animated proof for the formula giving the sum of the first integers 1+2+...+n.]]
To derive the above formula, begin by expressing the arithmetic series in two different ways:
:<math> S_n=a+a_2+a_3+\dots+a_{(n-1)} +a_n</math>
:<math> S_n=a+(a+d)+(a+2d)+\dots+(a+(n-2)d)+(a+(n-1)d). </math>
Rewriting the terms in reverse order:
:<math> S_n=(a+(n-1)d)+(a+(n-2)d)+\dots+(a+2d)+(a+d)+a.</math>

Adding the corresponding terms of both sides of the two equations and halving both sides:
:<math> S_n=\frac{n}{2}[2a + (n-1)d].</math>
This formula can be simplified as:
:<math>\begin{align}
S_n &=\frac{n}{2}[a + a + (n-1)d].\\
&=\frac{n}{2}(a+a_n).\\
&=\frac{n}{2}(\text{initial term}+\text{last term}).
\end{align}</math>
Furthermore, the mean value of the series can be calculated via: <math>S_n / n</math>:
:<math> \overline{a} =\frac{a_1 + a_n}{2}.</math>

The formula is essentially the same as the formula for the mean of a [[discrete uniform distribution]], interpreting the arithmetic progression as a set of equally probable outcomes.