# Arithmetic Sequences

How to solve the arithmetic sequence problems: formula, proof, examples, and their solutions.

## Formula

An arithmetic sequence is a sequence

whose differences of the adjacent terms

are the same (= *d*).

*a*_{n} = *a*_{1} + (*n* - 1)*d**a*_{n}: *n*th term*a*_{1}: first term*d*: common difference

## Proof

If the common difference is *d*, then*a*_{1} = *a*_{1}*a*_{2} = *a*_{1} + 1*d**a*_{3} = *a*_{1} + 2*d*

...*a*_{n} = *a*_{1} + (*n* - 1)*d*.

## Example 1

Find *a*_{1} and *d*.*a*_{1} = 1, *d* = +3

*a*_{1} = 1, *d* = +3*a*_{n} = 1 + (*n* - 1)⋅3

= 3*n* - 2

## Example 2

Find *a*_{1} and *d*.*a*_{1} = -2, *d* = +7

*a*_{1} = -2, *d* = +7*a*_{n} = -2 + (*n* - 1)⋅7

= 7*n* - 9

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

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

So *a*_{k} = 7*k* - 9 = 551.*k* = 80

## Example 3

Write *a*_{8} and *a*_{12}

by using *a*_{1} and *d*.*a*_{12} = *a*_{1} + 11*d* = 13*a*_{8} = *a*_{1} + 7*d* = 5

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

Use the elimination method to solve this system.

Then *d* = 2.

Elimination method

Put *d* = 2 into *a*_{1} + 7*d* = 5.

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

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

So *a*_{n} = -9 + (*n* - 1)⋅2

= 2*n* - 11