What Is the Arithmetic Sequence Formula?

An arithmetic sequence is a set of numbers in which each term after the first is obtained by adding a fixed constant called the common difference to the preceding term. This structure, fundamental in algebra and analysis, provides precise formulas for both the $n$-th term and the sum of $n$ terms, which are essential for systematically analyzing such patterns.

General Form and Element Structure of an Arithmetic Sequence

Let the first term of an arithmetic sequence be denoted by $a_1$, and let the common difference be $d$. The sequence can then be represented as:

$a_1,\ a_1 + d,\ a_1 + 2d,\ a_1 + 3d,\ldots,a_1 + (n-1)d$

For each integer $k$ $(1 \leq k \leq n)$, the $k$-th term is given by $a_k = a_1 + (k-1)d$.

Derivation of the $n$-th Term Formula for an Arithmetic Sequence

Let $a_1$ be the first term of the sequence and $d$ the common difference. To find the general term $a_n$:

The first term: $a_1$

The second term: $a_2 = a_1 + d$

The third term: $a_3 = a_2 + d$

Substitute $a_2$ into $a_3$:

$a_3 = (a_1 + d) + d = a_1 + 2d$

The fourth term: $a_4 = a_3 + d = a_1 + 2d + d = a_1 + 3d$

It follows by induction that the $n$-th term is:

$a_n = a_1 + (n-1)d$

$n$-th term formula: $a_n = a_1 + (n-1)d$

Alternatively, for any $k < n$, the $n$-th term in terms of $a_k$ is:

$a_n = a_k + (n - k)d$

Computation of the Common Difference in an Arithmetic Sequence

Given any two consecutive terms $a_{k}$ and $a_{k-1}$:

$a_k = a_{k-1} + d$

Subtract $a_{k-1}$ from both sides:

$a_k - a_{k-1} = d$

Common difference: $d = a_k - a_{k-1}$

For non-consecutive terms, if $a_m$ and $a_n$ are terms at positions $m$ and $n$ $(m > n)$,

$a_m = a_n + (m-n)d$

Isolate $d$:

$a_m - a_n = (m-n)d$

$d = \dfrac{a_m - a_n}{m - n}$

Explicit Derivation of the Sum of First $n$ Terms in an Arithmetic Sequence

Let $S_n$ denote the sum of the first $n$ terms:

$S_n = a_1 + a_2 + a_3 + \ldots + a_n$

Write the sequence forward and reverse, then sum term-wise:

$S_n = a_1 + (a_1 + d) + (a_1 + 2d) + \cdots + [a_1 + (n - 1)d]$

$S_n = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + \cdots + a_1$

Add these two equations together term-wise:

$2S_n = [a_1 + a_1 + (n-1)d] + [a_1 + d + a_1 + (n-2)d] + \ldots + [a_1 + (n-1)d + a_1]$

Each pair: $a_1 + [a_1 + (n-1)d] = 2a_1 + (n-1)d$

Thus, each of the $n$ pairs sums to $2a_1 + (n-1)d$.

There are $n$ such pairs, so:

$2S_n = n[2a_1 + (n-1)d]$

Divide both sides by $2$:

$S_n = \dfrac{n}{2}[2a_1 + (n-1)d]$

Sum formula: $S_n = \dfrac{n}{2}[2a_1 + (n-1)d]$

Since the $n$–th term is $a_n = a_1 + (n-1)d$, this formula can also be expressed as:

$S_n = \dfrac{n}{2}(a_1 + a_n)$

Explicit and Recursive Representation of Arithmetic Sequences

An arithmetic sequence can be specified either by an explicit formula or as a recurrence.

Explicit formula: $a_n = a_1 + (n-1)d$

Recursive formula: $\begin{cases} a_1 = a_1 \\ a_{n} = a_{n-1} + d\ \ \text{for}\ n \geq 2 \end{cases}$

Worked Example: Finding the $n$-th Term in a Given Arithmetic Sequence

Given: Sequence is $1,\,5,\,9,\,13,\ldots$ Find the $13$–th term.

First term: $a_1=1$

Common difference: $d=5-1=4$

Substituting into $a_n$: $a_{13} = 1 + (13-1)\cdot4$

$a_{13} = 1 + 12\cdot4$

$a_{13} = 1 + 48$

Final result: $a_{13}=49$

For sequences involving mixed progressions, refer to Arithmetic Geometric and Harmonic Progression.

Worked Example: Determining the First Term From a Given Term and Common Difference

Given: The $35$–th term is $687$; common difference $d=14$. Find $a_1$.

Given data: $a_{35}=687$, $d=14$, $n=35$

Substitution: $a_{35}=a_1+(35-1)14$

$687 = a_1 + (34)14$

$687 = a_1 + 476$

$a_1 = 687 - 476$

$a_1 = 211$

Final result: $a_1=211$

Worked Example: Calculating the Sum of Terms in an Arithmetic Sequence

Given: Find the sum of the series $3+7+11+\ldots$ up to $25$ terms.

First term: $a_1=3$

Common difference: $d=7-3=4$

Number of terms: $n=25$

First, compute $a_{25}$ using $a_{25} = a_1 + (25-1)d$:

$a_{25} = 3 + 24 \cdot 4$

$a_{25} = 3 + 96$

$a_{25} = 99$

Now, use the sum formula $S_{n} = \dfrac{n}{2}(a_1+a_n)$:

$S_{25} = \dfrac{25}{2}(3+99)$

$S_{25} = \dfrac{25}{2}(102)$

$S_{25} = 25 \times 51$

$S_{25} = 1275$

Final result: $S_{25}=1275$

Alternate Forms and Related Structures in Arithmetic Sequences

Expressing the $n$-th term in terms of a different known term $a_k$ gives $a_n = a_k + (n - k)d$, which facilitates variable entry points in sequence analysis. The common difference may also be found via the relation $d = \dfrac{a_n - a_k}{n - k}$ for non-consecutive terms. 

In understanding the distinction between sequences and series, as well as broader concepts relating arithmetic sequence formulas to pattern formation, see Difference Between Sequence and Series and Identifying Patterns.