- Home
- Precalculus
- Sequences and Series

# Arithmetic sequences

- Lesson: 1a3:09
- Lesson: 1b3:01
- Lesson: 1c7:05
- Lesson: 2a4:24
- Lesson: 2b5:46
- Lesson: 35:27

## What is a sequence?

A sequence is a list of things placed in a certain order. This lesson will teach you how to solve questions relating to sequences.

## What is an arithmetic sequence

When you come across an arithmetic sequence, the difference between one term and the next one is constant. So for example, when we have a sequence of $1, 2 , 3, 4, 5$, this is arithmetic since each number is simply +1 to the previous number. There’s actually a special name for the +$1$ in this situation…

## What is common difference

What is the common difference definition? When you figure out the constant difference between terms in an arithmetic sequence, you’ve found the common difference! In the previous example, $1$ is the common difference between the terms.

## How to find the common difference

When you’ve got an arithmetic sequence, how do you find the common difference? Firstly, your sequence should be in order. If you know that the sequence is arithmetic, then simply find the difference between the $2^{nd}$ and $1^{st}$ term and you’ll find the common difference. If you wanted to check your answer, feel free to find the difference between any consecutive terms and you should find that you’ll get the same answer for your common difference.

On the other hand, if you’re not sure if a sequence is arithmetic, but you find that there’s a constant difference between all the terms, you’ve proven that the sequence is arithmetic.

## How to find the $n^{th}$ term

When you are trying to find the $n^{th}$ term of an arithmetic sequence, use the following arithmetic sequence formula:

$t_{n}$ is the $n^{th}$ term, and $t_{1}$ is the first term. $d$ is the common difference in the arithmetic sequence. So if we use the example from above where we have a sequence of $1, 2 , 3, 4, 5$, if we had to find what would be the $6^{th}$ term to come after the number $5$ in our sequence, we’d get:

This makes sense since we know the next number to come should be $6$. Use the above formula and it will help you in finding the nth term. Let’s look at more examples on arithmetic sequences.

## Example problems

**Question 1:**

Arithmetic sequence formula

Consider the arithmetic sequence: 5, 9, 13, 17, …

a) Identify the common difference.

**Solution:**

Common difference is the difference between successive terms. We can pick any pair of successive terms to calculate the common difference. In this question, we have four terms:

$t_{1} = 5, t_{2} = 9, t_{3} = 13, t_{4} = 17$

We can pick any two successive terms from here. So:

We now know the common difference of this arithmetic sequence is $4$

b) Determine the seventh term of the sequence.

**Solution:**

We can use this equation to find the answer:

$t_{n} = t_{1} + (n - 1)d$

$t_{n} = n^{th}$ term

$t_{1} = 1^{st}$ term

$d$: common difference

We are looking for the seventh term. So:

$t_{n} = t_{1} + (n - 1)d$

$t_{7} = 5 + (7 - 1)4$

$t_{7} = 5 + 24$

$t_{7} = 29$

We can verify the answer by finding the terms up to the seventh term one by one. To do that, we just need to add the common difference to the last term:

$t_{4} = 17, t_{5} = 17 + 4 = 21, t_{6} = 21 + 4 = 25, t_{7} = 25 + 4 = 29$

c) Which term in the sequence has a value of $85$?

**Solution:**

We can use this equation to find the answer:

$t_{n} = t_{1} + (n - 1)d$

We know the $t_{n}, t_{1}$, and $d$. So we put the value into the equation

$85 = 5 + (n - 1)4$

Now we solve for $n$

$85 = 5 + (n - 1)4$

$80 = (n - 1)4$

$\frac{80}{4} = \frac{(n - 1)4}{4}$

$20 = n - 1$

$n = 21$

**Question 2:**

Determine $t_{1}, d, t_{n}$ for the sequences in which two terms are given

$t_{4} = 14, t_{10} = 32$

**Solution:**

We can use this equation as a starting point to find the answer:

$t_{n} = t_{1} + (n - 1)d$

We put the two terms we know to the equation:

$t_{4} = t_{1} + (4 - 1)d = 14 = t_{1} + 3d$

$t_{10} = t_{1} + (10 - 1)d = 32 = t_{1} + 9d$

Now we have two equations:

$14 = t_{1} + 3d$

$32 = t_{1} + 9d$

We can use elimination to solve for $d$ and $t_{1}$

$14 = t_{1} + 3d$

$(-) 32 = t_{1} + 9d$

$-18 = -6d$

$d = 3$

We put $d = 3$ into either one of the equations

$14 = t_{1} + 3(3)$

$14 = t_{1} + 9$

$t_{1} = 5$

We can find $t_{n}$ using the information we have

$t_{n} = t_{1} + (n - 1)d$

$t_{n} = 5 + (n - 1)3$

$t_{n} = 5 + 3n - 3$

$t_{n} = 3n + 2$

Want to double check your answers? Click here for a handy arithmetic sequence calculator.

Looking to move forward? Learn how to use pascal’s triangle or what the binomial theorem is.

Or expand on sequences and learn about monotonic and bounded sequences.

##### Do better in math today

##### Don't just watch, practice makes perfect.

### Arithmetic sequences

#### Lessons

• The nth term, ${t_n}$ ,of an arithmetic sequence:

${t_n} = {t_1} + \left( {n - 1} \right)d$

where, ${t_n}$: nth term

${t_1}$: first term

$d$ : common difference

- 1.
**Arithmetic sequence formula**

Consider the arithmetic sequence: 5, 9, 13, 17, … .a)Identify the common difference.b)Determine the seventh term of the sequence.c)Which term in the sequence has a value of 85? - 2.Determine $t_1,d,t_n$ for the sequences in which two terms are givena)$t_4=14$, $t_{10}=32$b)$t_3=-14$, $t_{12}=-59$
- 3.Three consecutive terms of an arithmetic sequence are written in the form:

$1+2x,7x,3+4x$

Solve for the value of x.

##### Do better in math today

##### Don't just watch, practice makes perfect.

### Arithmetic sequences

#### Don't just watch, practice makes perfect.

We have over 830 practice questions in Precalculus for you to master.

Get Started Now