Arithmetic sequence

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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

Example Problems and Solutions

Introductory Problems

Intermediate Problems

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

See Also