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# nth Term of an Arithmetic Sequence

If we know the first term (or a1) of an arithmetic sequence and the common difference d, we can plug that information into a formula to determine the nth term of the arithmetic sequence. This is called finding the general term. In this article, we'll go over what the formula is and how to apply it. Let's get started!

## Finding the nth term of an arithmetic sequence

The formula for finding the nth or general term of an arithmetic sequence is as follows:

${a}_{n}={a}_{1}+\left(n-1\right)d$

In this formula, n represents the term we're looking for, a1 is the first term in the sequence, and d is the common difference. This might make more sense with an example, so let's find the 27th term of the arithmetic sequence $5,8,11,14,...$

The first term is 5, so we have our ${a}_{1}$ . We weren't told what the common difference is, but we can figure it out by subtracting our ${a}_{1}$ from our ${a}_{2}$ (8):

$d=8-5=3$

We're looking for the 27th term of the sequence, allowing us to use the formula above to set up the following equation:

${a}_{27}=5+\left(27-1\right)3$

${a}_{27}=5+26\left(3\right)$

${a}_{27}=5+78$

${a}_{27}=83$

As long as we remember to plug in the correct values, the math involved is relatively straightforward. Let's try another example: Find the 40th term of an arithmetic sequence in which ${a}_{8}=60$ and ${a}_{12}=48$ .

This one is a little more complicated since our formula requires the first term ( ${a}_{1}$ ) and the common difference and we weren't provided with either. However, we can set up a system of linear equations using our formula and the information we were given to solve the problem. Let's start by writing our formula:

${a}_{n}={a}_{1}+\left(n-1\right)d$

Next, we substitute 60 for ${a}_{8}$ and 48 for ${a}_{12}$

$60={a}_{1}+\left(8-1\right)d={a}_{1}+7d$

$48={a}_{1}+\left(12-1\right)d={a}_{1}+11d$

Now, we can subtract the second equation from the first equation and solve for d:

${a}_{8}-{a}_{12}=-4d$

$60-48=12=-4d$

$d=-3$

Now that we know the common difference, it's time to figure out ${a}_{1}$ . We can do so using either of the terms we were given, so let's use ${a}_{8}$ :

$60={a}_{1}+7\left(-3\right)$

$60={a}_{1}-21$

$81={a}_{1}$

We have values for both d and ${a}_{1}$ now, so we can finally use our formula to determine the 40th term in the sequence:

${a}_{40}=81+39\left(-3\right)=81-117=-36$

Our final answer is -36. There were a lot of steps involved, and it's important to proceed carefully to avoid making any unnecessary errors.

Depending on the pattern, we may be able to eyeball the pattern and work it out one term at a time. This is frequently more time-consuming than using the formula, however, so watch the clock!

## Practice questions on finding the nth term of an arithmetic sequence

a. Find the 15th term of the following arithmetic sequence: $0,2,4,6,8,...$

Whenever we're looking for the general term of an arithmetic sequence, our first move should always be to write our formula:

${a}_{n}={a}_{1}+\left(n-1\right)d$

Our ${a}_{1}$ value is 0 and our ${a}_{2}$ value is 2, so we can determine the common difference (d) by subtracting them:

$d=2-0=2$

We now have all of the values we need to determine the 15th term of the sequence using our formula:

${a}_{15}=0+14\left(2\right)$

${a}_{15}=28$

b. If the first day of the year is Monday, what day of the week will the 295th day be?

It might not seem like it, but this is an arithmetic sequence question. Our formula is:

${a}_{n}={a}_{1}+\left(n-1\right)d$

${a}_{295}={a}_{1}+\left(295-1\right)d$

${a}_{295}=294d$

That math looks like a nightmare, but fortunately, we won't have to do it since we're looking for a day of the week. There are 7 days per week, so we divide 294 days by 7 to determine how many weeks have passed. The answer is 42 with no remainder, which means we're back on the same day we started with: Monday.

## Flashcards covering the nth Term of an Arithmetic Sequence

Algebra 1 Flashcards

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