# Arithmetic Sequence

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/arithmetic-sequence/
Fetched from algebrica.org post 14634; source modified 2026-03-12T20:56:26.

## What is an arithmetic sequence

A [sequence](../sequences.md) \\( a\_n \\) is called an **arithmetic sequence** (or arithmetic progression) if it consists of numbers arranged in such a way that the difference between any term and the one before it is constant. It is characterized by terms of the form:

\\[
a\_1, a\_2, \\dots, a\_n \\quad \\text{with} \\quad a\_n - a\_{n-1} = d
\\]

-   By convention, the first term of an arithmetic progression is typically indexed with \\( n = 1.\\)
-   \\( d \\) represents the difference between two consecutive terms in an arithmetic progression, and it is known as the common difference.
-   If \\( d > 0 \\), the progression is increasing.
-   If \\( d < 0 \\), the progression is decreasing.
-   If \\( d = 0 \\), the progression is constant.

Let’s consider, for example, the sequence of non-negative even numbers:

![](https://algebrica.org/wp-content/uploads/resources/images/arithmetic-sequence.png)

* * *

An arithmetic sequence can also be defined using a recursive formula:

\\[
a\_n = a\_1 + n \\cdot d \\quad \\text{where } a\_1, d \\in \\mathbb{R}
\\]

![](https://algebrica.org/wp-content/uploads/resources/images/arithmetic-sequence-2.png)

An arithmetic progression exhibits a characteristic stepwise pattern, where the height of each step corresponds to the common difference between consecutive terms in the sequence.

* * *

In an arithmetic progression, each term \\( a\_n \\) is obtained by adding the first term \\( a\_1 \\) to the product of the common difference \\( d \\) and \\( (n - 1) \\). This gives the general formula for the \\( n \\)-th term:

\\[
a\_n = a\_1 + (n - 1) \\cdot d \\quad \\text{for } n \\geq 1
\\]

This formula allows you to compute any term in the sequence directly, without listing all the previous ones.

## Example

Let’s define an arithmetic sequence with first term \\( a\_1 = 2 \\) and common difference \\( d = 3 \\). We use the formula:

\\[
a\_n = a\_1 + (n - 1) \\cdot d
\\]

Plug in the values:

\\[
a\_n = 2 + (n - 1) \\cdot 3
\\]

Now calculate the first few terms:

-   \\( a\_1 = 2 \\)
-   \\( a\_2 = 2 + 1 \\cdot 3 = 5 \\)
-   \\( a\_3 = 2 + 2 \\cdot 3 = 8 \\)
-   \\( a\_4 = 2 + 3 \\cdot 3 = 11 \\)
-   \\( a\_5 = 2 + 4 \\cdot 3 = 14 \\)

The resulting sequence is:

\\[
2,\\ 5,\\ 8,\\ 11,\\ 14,\\ \\dots
\\]

## Sum of \\(n\\) terms of an arithmetic progression

The sum \\( S\_n \\) of the first \\( n \\) terms \\( a\_1, a\_2, \\dots, a\_n \\) of an arithmetic progression is equal to the product of \\( n \\) and the average of the first and last term:

\\[
S\_n = n \\cdot \\frac{a\_1 + a\_n}{2}
\\]

This formula allows you to quickly compute the total sum of a finite number of terms in an arithmetic progression. For example, consider the arithmetic progression of non-negative even numbers:

\\[
2,\\ 4,\\ 6,\\ 8,\\ 10
\\]

We want to calculate the sum of the first 5 terms \\( (n = 5).\\) Using the formula, we have:

\\[
S\_5 = 5 \\cdot \\frac{2 + 10}{2} = 5 \\cdot 6 = 30
\\]

##### This illustrates the same reasoning behind Gauss’s trick: by pairing the first and last terms, you can quickly compute the total sum of an arithmetic progression.