Difference between revisions of "Arithmetic sequence"
(→Arithmetic Sequence Video) |
Etmetalakret (talk | contribs) |
||
Line 1: | Line 1: | ||
− | + | In [[algebra]], an '''arithmetic sequence''', sometimes called an '''arithmetic progression''', is a [[sequence]] of numbers such that the difference between any two consecutive terms is constant. This constant is called the '''common difference''' of the sequence. | |
− | [ | ||
− | + | For example, <math>-7, 0, 7, 14</math> is an arithmetic sequence with common difference <math>7</math> and <math>99, 91, 83, 75, \ldots</math> is an arithmetic sequence with common difference <math>-8</math>; However, <math>1, 2, 3, -4</math> and <math>4, 12, 36, 108</math> are not arithmetic sequences, as the difference between consecutive terms varies. | |
− | |||
− | + | More formally, the sequence <math>a_1, a_2, \ldots , a_n</math> is an arithmetic progression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. This definition appears most frequently in its three-term form; that constants <math>a</math>, <math>b</math>, and <math>c</math> are in arithmetic progression if and only if <math>b - a = c - b</math>. | |
− | |||
− | < | ||
− | |||
− | == | + | == Properties == |
− | + | Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let <math>a_1</math> be the first term, <math>a_n</math> be the <math>n</math>th term, and <math>d</math> be the common difference of any arithmetic sequence; then, <math>a_n = a_1 + (n-1)d</math>. | |
− | |||
− | <math> | + | A common lemma is that given the <math>n</math>th term <math>x</math> and <math>m</math>th term <math>y</math> of an arithmetic sequence, the common difference is equal to <math>\frac{y-x}{m-n}</math>. |
− | + | ''Proof'': Let the sequence have first term <math>a_1</math> and common difference <math>d</math>. Then using the above result, <cmath>\frac{y-x}{m-n} = \frac{(a_1 + (m - 1)d) - (a_1 + (n-1)d)}{m-n} = \frac{dm - dn}{m-n} = d,</cmath> as desired. <math>\square</math> | |
− | <math>\frac{ | + | Another lemma is that for any consecutive terms <math>a_{n-1}</math>, <math>a_n</math>, and <math>a_{n+1}</math> of an arithmetic sequence, then <math>a_n</math> is the average of <math>a_{n-1}</math> and <math>a_{n+1}</math>. In symbols, <math>a_n = \frac{a_{n-1} + a_{n+1}}{2}</math>. This is mostly used to perform substitutions. |
+ | |||
+ | == Sum == | ||
== Example Problems and Solutions == | == Example Problems and Solutions == |
Revision as of 21:29, 29 August 2021
In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence.
For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; However, and are not arithmetic sequences, as the difference between consecutive terms varies.
More formally, the sequence is an arithmetic progression if and only if . This definition appears most frequently in its three-term form; that constants , , and are in arithmetic progression if and only if .
Contents
[hide]Properties
Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, .
A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to .
Proof: Let the sequence have first term and common difference . Then using the above result, as desired.
Another lemma is that for any consecutive terms , , and of an arithmetic sequence, then is the average of and . In symbols, . This is mostly used to perform substitutions.
Sum
Example Problems and Solutions
Introductory Problems
- 2005 AMC 10A Problem 17
- 2006 AMC 10A Problem 19
- 2012 AIME I Problems/Problem 2
- 2004 AMC 10B Problems/Problem 10
Intermediate Problems
- Find the roots of the polynomial , given that the roots form an arithmetic progression.