|
|
| (5 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | An '''arithmetic series''' is a sum of consecutive terms in an [[arithmetic sequence]]. For instance,
| + | #REDIRECT[[Arithmetic sequence]] |
| − | | |
| − | <math> 2 + 6 + 10 + 14 + 18 </math>
| |
| − | | |
| − | 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):
| |
| − | <cmath>\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*}</cmath>
| |
| − | | |
| − | Now, adding vertically and shifted over one, we get
| |
| − | | |
| − | <cmath>2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)</cmath>
| |
| − | | |
| − | This equals <math>2S = n(2a+(n-1)d)</math>, so the sum is <math>\frac{n}{2} (2a+(n-1)d)</math>.
| |
| − | | |
| − | == Problems ==
| |
| − | === Introductory Problems ===
| |
| − | * [[2006_AMC_10A_Problems/Problem_9 | 2006 AMC 10A, Problem 9]]
| |
| − | *[[2006 AMC 12A Problems/Problem 12 | 2006 AMC 12A, Problem 12]]
| |
| − | | |
| − | === Intermediate Problems ===
| |
| − | *[[2003 AIME I Problems/Problem 2|2003 AIME I, Problem 2]]
| |
| − | | |
| − | === Olympiad Problem ===
| |
| − | | |
| − | == See also ==
| |
| − | * [[Series]]
| |
| − | * [[Summation]]
| |
| − | | |
| − | {{stub}}
| |
