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 17: / Line 17: @@
 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 fist 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>.
+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 + (n-1)d}{2}</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>
+'''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 ==

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