# Find a formula for {eq}a_n {/eq} for the arithmetic sequence. {eq}a_1 = 15 {/eq}, {eq}d = 4 {/eq}

## Question:

Find a formula for {eq}a_n {/eq} for the arithmetic sequence.

{eq}a_1 = 15 {/eq}, {eq}d = 4 {/eq}

### The {eq}n^{\text{th}} {/eq} term formula for the Arithmetic Sequence:

In an arithmetic sequence, the {eq}n^{\text{th}} {/eq} term is denoted as {eq}a_n. {/eq} For instance, the first term is denoted as {eq}a_1. {/eq} The difference between each consecutive term in the sequence is indicated as {eq}d, {/eq} which is a constant. Hence, the formula for the {eq}n^{\text{th}} {/eq} term of the arithmetic sequence will be:

$$a_n = a+ \left(n-1 \right) d$$

We have,

• The first term, {eq}a_1 = 15 {/eq}
• Common difference: {eq}d = 4 {/eq}

The {eq}n^{\text{th}} {/eq} term is, {eq}a_n = a+ \left(n-1 \right) d {/eq}.

Let us apply the given values in the above {eq}n^{\text{th}} {/eq} formula:

\begin{align*} a_n &= a+ \left(n-1 \right) d \\ \\ &= 15 + \left( n -1 \right) \times 4 \\ \\ &= 15 + 4n - 4 \\ \\ \therefore a_n &= 11 + 4n \end{align*}

Hence, the {eq}n^{\text{th}} {/eq} formula is {eq}\color{blue}{a_n = 11+4n} {/eq}.