Difference between revisions of "Arithmetic sequence"

Latest revision as of 01:00, 7 March 2024

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, $1, 2, 3, 4$ is an arithmetic sequence with common difference $1$ and $99, 91, 83, 75$ is an arithmetic sequence with common difference $-8$ ; however, $7, 0, 7, 14$ and $4, 12, 36, 108, \ldots$ are not arithmetic sequences, as the difference between consecutive terms varies.

More formally, the sequence $a_1, a_2, \ldots , a_n$ is an arithmetic progression if and only if $a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}$ . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants $a$ , $b$ , and $c$ are in arithmetic progression if and only if $b - a = c - b$ .

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 $a_1$ be the first term, $a_n$ be the $n$ th term, and $d$ be the common difference of any arithmetic sequence; then, $a_n = a_1 + (n-1)d$ .

A common lemma is that given the $n$ th term $x$ and $m$ th term $y$ of an arithmetic sequence, the common difference is equal to $\frac{y-x}{m-n}$ .

Proof: Let the sequence have first term $a_1$ and common difference $d$ . Then using the above result, $\frac{y-x}{m-n} = \frac{(a_1 + (m - 1)d) - (a_1 + (n-1)d)}{m-n} = \frac{dm - dn}{m-n} = d,$ as desired. $\square$

Another common lemma is that a sequence is in arithmetic progression if and only if $a_n$ is the arithmetic mean of $a_{n-1}$ and $a_{n+1}$ for any consecutive terms $a_{n-1}, a_n, a_{n+1}$ . In symbols, $a_n = \frac{a_{n-1} + a_{n+1}}{2}$ . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences.

Sum

An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value.

The first is that if an arithmetic series has first term $a_1$ , last term $a_n$ , and $n$ total terms, then its value is equal to $\frac{n(a_1 + a_n)}{2}$ .

Proof: Let the series be equal to $S$ , and let its common difference be $d$ . Then, we can write $S$ in two ways: $S = a_1 + (a_1 + d) + \cdots + (a_1 + (n-1)d)$ $S = a_n + (a_n - d) + \cdots + (a_n - (n-1)d.$ Adding these two equations cancels all terms involving $d$ ; $2S = (a_1 + a_n) + (a_1 + a_n) + \cdots + (a_1 + a_n) = n(a_1 + a_n),$ and so $S = \frac{n(a_1 + a_n)}{2}$ , as required. $\square$

The second is that if an arithmetic series has first term $a_1$ , common difference $d$ , and $n$ terms, it has value $\frac{n(2a_1 + (n-1)d)}{2}$ .

Proof: The final term has value $a_1 + (n-1)d$ . Then by the above formula, the series has value $\frac{n(a_1 + (a_1 + (n-1)d)}{2} = \frac{n(2a_1 + (n-1)d)}{2}.$ This completes the proof. $\square$

Problems

Here are some problems with solutions that utilize arithmetic sequences and series.

Introductory problems

Intermediate problems

2003 AIME I, Problem 2
Find the roots of the polynomial $x^5-5x^4-35x^3+ax^2+bx+c$ , given that the roots form an arithmetic progression.

@@ Line 1: / Line 1: @@
-==Definition==
+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.
-An '''arithmetic sequence''' is a [[sequence]] of numbers in which each term is given by adding a fixed value to the previous term.  For example, -2, 1, 4, 7, 10, ... is an arithmetic sequence because each term is three more than the previous term. In this case, 3 is called the ''common difference'' of the sequence.  More formally, an arithmetic sequence <math>a_n</math> is defined [[recursion|recursively]] by a first term <math>a_0</math> and <math>a_n = a_{n-1} + d</math> for <math>n \geq 1</math>, where <math>d</math> is the common difference. Explicitly, it can be defined as <math>a_n=a_0+dn</math>.
-==Sums of Arithmetic Sequences==
+For example, <math>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math> and <math>99, 91, 83, 75</math> is an arithmetic sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference between consecutive terms varies.
-There are many ways of calculating the sum of the terms of a [[finite]] arithmetic sequence.  Perhaps the simplest is to take the average, or [[arithmetic mean]], of the first and last term and to multiply this by the number of terms. Formally, <math>s_n=\frac{n}{2}(a_1+a_n)</math>. For example,
+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>. A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, 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>.
-<math>\displaystyle 5 + 7 + 9 + 11 + 13 + 15 + 17 = \frac{5+17}{2} \cdot 7 = 77</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>.
-or
+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>.
-<math>\frac{7}{2}(5+17)=77</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>
-== Example Problems and Solutions ==
+Another common lemma is that a sequence is in arithmetic progression if and only if <math>a_n</math> is the [[arithmetic mean]] of <math>a_{n-1}</math> and <math>a_{n+1}</math> for any consecutive terms <math>a_{n-1}, a_n, 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, though it occasionally serves as a definition of arithmetic sequences.
-=== Introductory Problems ===
+== Sum ==
+An '''arithmetic series''' is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value.
+The first is that if an arithmetic series has first term <math>a_1</math>, last term <math>a_n</math>, and <math>n</math> total terms, then its value is equal to <math>\frac{n(a_1 + a_n)}{2}</math>.
+'''Proof''': Let the series be equal to <math>S</math>, and let its common difference be <math>d</math>. Then, we can write <math>S</math> in two ways: <cmath>S = a_1 + (a_1 + d) + \cdots + (a_1 + (n-1)d)</cmath> <cmath>S = a_n + (a_n - d) + \cdots + (a_n - (n-1)d.</cmath> Adding these two equations cancels all terms involving <math>d</math>; <cmath>2S = (a_1 + a_n) + (a_1 + a_n) + \cdots + (a_1 + a_n) = n(a_1 + a_n),</cmath> and so <math>S = \frac{n(a_1 + a_n)}{2}</math>, as required. <math>\square</math>
+The second is that if an arithmetic series has first term <math>a_1</math>, common difference <math>d</math>, and <math>n</math> terms, it has value <math>\frac{n(2a_1 + (n-1)d)}{2}</math>.
+'''Proof''': The final term has value <math>a_1 + (n-1)d</math>. Then by the above formula, the series has value <cmath>\frac{n(a_1 + (a_1 + (n-1)d)}{2} = \frac{n(2a_1 + (n-1)d)}{2}.</cmath> This completes the proof. <math>\square</math>
+== Problems ==
+Here are some problems with solutions that utilize arithmetic sequences and series.
+=== Introductory problems ===
 * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]
 * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]]
 * [[2012 AIME I Problems/Problem 2]]
+* [[2004 AMC 10B Problems/Problem 10]]
+* [[2006_AMC_10A_Problems/Problem_9 | 2006 AMC 10A, Problem 9]]
+* [[2006 AMC 12A Problems/Problem 12 | 2006 AMC 12A, Problem 12]]
-=== Intermediate Problems ===
+=== Intermediate problems ===
+* [[2003 AIME I Problems/Problem 2|2003 AIME I, Problem 2]]
 * Find the roots of the polynomial <math>x^5-5x^4-35x^3+ax^2+bx+c</math>, given that the roots form an arithmetic progression.
-==See Also==
+== See Also ==
-*[[sequence|Sequence]]
+* [[Geometric sequence]]
-*[[series|Series]]
+* [[Harmonic sequence]]
-*[[geometric sequence|Geometric Sequences]]
+* [[Sequence]]
+* [[Series]]
+[[Category:Algebra]] [[Category:Sequences and series]] [[Category:Definition]]

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