# Arithmetic Series

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

## Formula

An arithmetic series is the sum

of the arithmetic sequences' terms.

There are two formulas to find the sum:*S*_{n} = (*n*/2)[2*a*_{1} + (*n* - 1)*d*]

= (*n*/2)[*a*_{1} + *a*_{n}]*S*_{n}: the sum of '1st term ~ *n*th term'*a*_{1}: first term*d*: common difference*a*_{n}: *n*th term

## Proof

*S*_{n} = *a*_{1} + *a*_{2} + *a*_{3} + ... + *a*_{n}

= *a*_{1} + [*a*_{1} + 1*d*] + [*a*_{1} + 2*d*] + ... + [*a*_{1} + (*n* - 1)*d*]

Arithmetic sequences

Write the same *S*_{n} in the opposite order.*S*_{n} = *a*_{n} + *a*_{n - 1} + *a*_{n - 2} + ... + *a*_{1}

= [*a*_{1} + (*n* - 1)*d*] + [*a*_{1} + (*n* - 2)*d*] + [*a*_{1} + (*n* - 3)*d*] + ... + *a*_{1}

Add these two.

Then (left side) = 2*S*_{n}.

In the right side, each term is [2*a*_{1} + (*n* - 1)*d*].

And there are *n* terms.

So (right side) = *n*⋅[2*a*_{1} + (*n* - 1)*d*].

Divide both sides by 2.

Then *S*_{n} = (*n*/2)[2*a*_{1} + (*n* - 1)*d*].

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

= *a*_{1} + *a*_{n}

So *S*_{n} = (*n*/2)⋅[*a*_{1} + *a*_{n}].

## Example 1

*a*_{1} = 3, *d* = 5, *d* = 20

Use the first formula.*S*_{20} = (20/2)[3 + (20 - 1)⋅5]

## Example 2

*a*_{1} = -1, *d* = +3

So *a*_{n} = -1 + (*n* - 1)⋅4.

Arithmetic sequences

*a*_{n} = 4*n* - 5

And the last term, *a*_{n}, is 31.

So 4*n* - 5 = 31.*n* = 9 (number of the terms)

*n* = 9, *a*_{1} = -1, *a*_{n} = 31

Use the second formula.*S*_{9} = (9/2)[-1 + 31]

## Example 3

Case 1: *n* = 2, 3, 4, ...*S*_{n} = *a*_{1} + *a*_{2} + ... + *a*_{n - 1} + *a*_{n}*S*_{n - 1} = *a*_{1} + *a*_{2} + ... + *a*_{n - 1}

Subtract these two.*S*_{n} - *S*_{n - 1} = *a*_{n}

So use *a*_{n} = *S*_{n} - *S*_{n - 1}

to find *a*_{n}.*a*_{n} = 2*n* + 1

(This is true when *n* = 2, 3, 4, ... .)

Case 2: *n* = 1

See if *a*_{n} = 2*n* + 1 is true when *n* = 1

by checking *a*_{1} = *S*_{1}.*a*_{1} = *S*_{1} = 3

So *a*_{n} = 2*n* + 1 is true when *n* = 1.

So *a*_{n} = 2*n* + 1 for all *n*.

(*n* = 1, 2, 3, 4, ...)