Arithmetic series: Difference between revisions
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is an arithmetic series whose value is 50. | 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): | |||
S = a + (a+d) + (a+2d) + ... + (a+(n-1)d) | |||
S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a | |||
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 <math>\displaystyle \frac{n}{2} (2a+(n-1)d</math> | |||
== Example Problems == | == Example Problems == | ||
Revision as of 20:40, 6 August 2006
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An arithmetic series is a sum of consecutive terms in an arithmetic sequence. For instance,
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): S = a + (a+d) + (a+2d) + ... + (a+(n-1)d)
S = (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a
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 
Example Problems
Introductory Problems
