# Arithmetic Sequences

How to find the *n*th term of an arithmetic sequence: formula, examples, and their solutions.

## Formula

An arithmetic sequence is a sequence

whose differences of the adjacent terms

are the same (= *d*).*a*_{2} - *a*_{1} = *d**a*_{3} - *a*_{2} = *d**a*_{4} - *a*_{3} = *d*

...*d* is the [common difference].

So to find the next term,

add the common difference: +*d*.

The *n*th term of an arithmetic sequence, *a*_{n},

can be found by using the formula below.*a*_{n} = *a*_{1} + (*n* - 1)*d**a*_{n}: *n*th term*a*_{1}: First term*d*: Common difference

## Example 1

The first term is 1.

So *a*_{1} = 1.

4 - 1 = 3

7 - 4 = 3

10 - 7 = 3

13 - 10 = 3

...

So *d* = 3.

So *a*_{n} = 1 + (*n* - 1)⋅3.

+(*n* - 1)⋅3 = +3*n* - 3

1 - 3 = -2

So *a*_{n} = 3*n* - 2.

## Example 2

The first term is -2.

So *a*_{1} = -2.

5 - (-2) = 7

12 - 5 = 7

19 - 12 = 7

...

So *d* = 7.

So *a*_{k} = -2 + (*k* - 1)⋅7.

+(*k* - 1)⋅7 = +7*k* - 7

-2 - 7 = -9

So *a*_{k} = 7*k* - 9.

*a*_{k} = 7*k* - 9

And it says*a*_{k} = 551.

So 7*k* - 9 = 551.

Move -9 to the right side.

Then 7*k* = 560.

Divide both sides by 7.

Then *k* = 80.

## Example 3

*a*_{8} = *a*_{1} + 7*d* = 5*a*_{12} = *a*_{1} + 11*d* = 13

So*a*_{1} + 7*d* = 5,*a*_{1} + 11*d* = 13.

The goal is to find *a*_{1} and *d*.

*a*_{1} + 11*d* = 13*a*_{1} + 7*d* = 5

Subtract these two equations

to eliminate *a*_{1}.

Then 4*d* = 8.

Elimination method

Divide both sides by 4.

Then *d* = 2.

Put [*d* = 2]

into [*a*_{1} + 7*d* = 5].

Then *a*_{1} + 7⋅2 = 5.

+7⋅2 = +14

Move +14 to the right side.

Then *a*_{1} = -9.

*a*_{1} = -9*d* = 2

So *a*_{n} = -9 + (*n* - 1)⋅2.

+(*n* - 1)⋅2 = +2*n* - 2

-9 - 2 = -11

So *a*_{n} = 2*n* - 11.