Difference between revisions of "Arithmetic sequence"

Revision as of 15:38, 3 September 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, $-7, 0, 7, 14$ is an arithmetic sequence with common difference $7$ and $99, 91, 83, 75, \ldots$ is an arithmetic sequence with common difference $-8$ ; However, $1, 2, 3, -4$ and $4, 12, 36, 108$ 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 $a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}$ . This definition appears most frequently in its three-term form; 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 lemma is that for any consecutive terms $a_{n-1}$ , $a_n$ , and $a_{n+1}$ of an arithmetic sequence, then $a_n$ is the average of $a_{n-1}$ and $a_{n+1}$ . In symbols, $a_n = \frac{a_{n-1} + a_{n+1}}{2}$ . This is mostly used to perform substitutions.

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

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 that test knowledge of 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 10: / Line 10: @@
 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>
+'''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>
 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.
@@ Line 19: / Line 19: @@
 The fist is that if an arithmetic sequence 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 sequence 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>
+'''Proof''': Let the sequence 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 sequence 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>.
-''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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