Difference between revisions of "Arithmetic series"

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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):
<div style="text-align:center;"><math>\displaystyle S =  a + (a+d) + (a+2d) + ... + (a+(n-1)d)</math>
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<cmath>\begin{align*}
 
+
S &=  a + (a+d) + (a+2d) + ... + (a+(n-1)d) \
<math>\displaystyle S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a</math></div>
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S &= (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a
 +
\end{align*}</cmath>
  
 
Now, adding vertically and shifted over one, we get
 
Now, adding vertically and shifted over one, we get
  
<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>
+
<cmath>2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)</cmath>
  
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>.
+
This equals <math>2S = n(2a+(n-1)d)</math>, so the sum is <math>\frac{n}{2} (2a+(n-1)d)</math>.
  
 
== Problems ==
 
== Problems ==

Revision as of 13:31, 24 April 2009

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): \begin{align*} S &=  a + (a+d) + (a+2d) + ... + (a+(n-1)d) \\ S &= (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a \end{align*}

Now, adding vertically and shifted over one, we get

\[2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)\]

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

Problems

Introductory Problems

Intermediate Problems

Olympiad Problem

See also

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