Arithmetic Progression

Consider the following sequences:

2, 7, 12, 17, 22, 27, \ldots; \quad 1;\; 1.5;\; 2;\; 2.5;\; 3;\; 3.5;\; \ldots; \quad 3, 1, -1, -3, -5, -7, \ldots

They share a characteristic feature: each subsequent term is obtained by adding the same number to the previous term. For the first sequence this number is 5, for the second it is 0.5, and for the third it is $-2$ . Such sequences are called arithmetic progressions (from Latin progressio — forward movement).

Definition

📐Definition — Arithmetic Progression

An arithmetic progression is a sequence in which each term, starting from the second, equals the previous term plus the same constant number.

This constant number equals the difference between any consecutive terms of the sequence. It is called the common difference of the arithmetic progression and is denoted by $d$ (from the Latin word differentia — difference).

If $(a_n)$ is an arithmetic progression with common difference $d$ , then

d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \ldots,

that is, for any natural number $n$ we have $a_{n+1} - a_n = d$ . This gives the recurrent formula:

a_{n+1} = a_n + d.

ℹNote — Defining an Arithmetic Progression Recurrently

An arithmetic progression can be defined recurrently:

a_1 = a, \quad a_{n+1} = a_n + d.

Thus, to define an arithmetic progression, one needs to specify its first term and common difference.

Some examples:

If $a_1 = 2$ and $d = 5$ : $2, 7, 12, 17, 22, 27, \ldots$
If $a_1 = 1$ and $d = 2$ , we get the odd numbers: $1, 3, 5, 7, 9, 11, \ldots$
A stationary sequence is an arithmetic progression with $d = 0$ .

Formula for the $n$ -th Term

From the definition of an arithmetic progression $(a_n)$ it follows:

a_2 = a_1 + d, \quad a_3 = a_1 + 2d, \quad a_4 = a_1 + 3d, \quad a_5 = a_1 + 4d.

These examples lead to the following conclusion:

⚡Theorem — Formula for the n-th Term of an Arithmetic Progression

a_n = a_1 + d(n - 1)

This can be proved by mathematical induction.

Consider a function $f(x) = kx + b$ with $D(f) = \mathbb{N}$ or $D(f) = \{1, 2, \ldots, n\}$ . It forms an arithmetic progression with common difference $k$ . Indeed, $f(n+1) - f(n) = k(n+1) + b - (kn + b) = k$ . This means the sequence $f(1), f(2), \ldots, f(n), \ldots$ is an arithmetic progression with common difference $k$ .

Property of Terms of an Arithmetic Progression

⚡Theorem — Characteristic Property

Any term of an arithmetic progression, except the first, equals the arithmetic mean of its two neighboring terms. That is, for $n > 1$ :

a_n = \frac{a_{n-1} + a_{n+1}}{2}

The converse is also true: if in a sequence every term except the first (and last, if the progression is finite) equals the arithmetic mean of its two neighbors, then this sequence is an arithmetic progression.

⚡Theorem — Theorem 31.1

A sequence with more than two terms is an arithmetic progression if and only if every term, except the first (and last, if the sequence is finite), equals the arithmetic mean of its two neighboring terms.

Examples

✎Example — Finding the First Term and Common Difference

Problem. The terms of an arithmetic progression $(a_n)$ are integers. It is known that $a_3 \cdot a_6 = 406$ and when the ninth term is divided by the fourth term, the quotient is 2 with remainder 6. Find the first term and the common difference.

Solution. Write: $a_3 = a_1 + 2d$ , $a_4 = a_1 + 3d$ , $a_6 = a_1 + 5d$ , $a_9 = a_1 + 8d$ . Using the given conditions, we form the system:

\begin{cases} (a_1 + 2d)(a_1 + 5d) = 406, \\ a_1 + 8d = 2(a_1 + 3d) + 6. \end{cases}

Solving this system yields:

\begin{bmatrix} a_1 = 4, \\ d = 5; \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} a_1 = -\dfrac{79}{7}, \\ d = -\dfrac{37}{14}. \end{bmatrix}

Since the terms must be integers, the answer is $a_1 = 4$ , $d = 5$ .

Answer: $a_1 = 4$ , $d = 5$ .

✎Example — Irrational Numbers and Arithmetic Progression

Problem. Can the numbers $1$ , $\sqrt{2}$ , $\sqrt{3}$ be terms of the same arithmetic progression?

Solution. Suppose $1$ , $\sqrt{2}$ , $\sqrt{3}$ are terms of an arithmetic progression $(a_n)$ with common difference $d$ ( $d \neq 0$ ), corresponding to $a_n$ , $a_m$ , $a_k$ respectively. Then:

1 = a_1 + d(n-1), \quad \sqrt{2} = a_1 + d(m-1), \quad \sqrt{3} = a_1 + d(k-1).

This gives $\sqrt{2} - 1 = d(m - n)$ and $\sqrt{3} - 1 = d(k - n)$ .

Since $k \neq n$ , we get $\dfrac{\sqrt{2} - 1}{\sqrt{3} - 1} = \dfrac{m - n}{k - n}$ .

The right-hand side is rational. It can be shown that the left-hand side is irrational. Therefore, the equality is false.

Thus, $1$ , $\sqrt{2}$ , $\sqrt{3}$ cannot be terms of the same arithmetic progression.

Answer: no.

Exercises

✏Exercise — Arithmetic Progression Problems

31.3. The first term of an arithmetic progression is $a_1 = 4$ , and the common difference is $d = 0.4$ . Find: 1) $a_5$ ; 2) $a_{11}$ ; 3) $a_{32}$ .

31.8. Find the formula for the $n$ -th term of the arithmetic progression:

$-5, -7, -9, -11, \ldots$
$a^2, 2a^2, 3a^2, 4a^2, \ldots$

31.9. Is the following number a term of the arithmetic progression $(c_n)$ :

The number 20.4, if $c_1 = 11.4$ and $d = 0.6$ ?
The number 38, if $c_1 = 8$ and $d = 1.4$ ?

31.15. Is the sequence $(a_n)$ an arithmetic progression if it is defined by:

$a_n = -6n + 3$
$a_n = 2n^2 - n$
$a_n = -2.8n$