# Arithmetic Series

How to find the value of an arithmetic series: formulas, examples, and their solutions.

## Formulas

A series means the sum of the sequence.

So an arithmetic series is

the sum of the terms of the arithmetic sequence.

Arithmetic sequences*S*_{n} is the symbol of a series:*S*_{n} = *a*_{1} + *a*_{2} + *a*_{3} + ... + *a*_{n}.

For an arithmetic series *S*_{n},*S*_{n} = (*n*/2)[2*a*_{1} + (*n* - 1)*d*].*a*_{1}: First term*d*: Common difference*a*_{n}: *n*th term

If *a*_{n} is given,

then you can also use this formula.*S*_{n} = (*n*/2)[*a*_{1} + *a*_{n}]*a*_{n}: *n*th term

## Example 1

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

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

20/2 = 10

20 - 1 = 19

19⋅5 = 95

6 + 95 = 101

10⋅101 = 1010

So *S*_{20} = 1010.

## Example 2

The first term is -1.

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

7 - 3 = 4

So *d* = 4.

So *a*_{n} = -1 + (*n* - 1)⋅4.

Arithmetic sequences

(*n* - 1)⋅4 = 4*n* - 4

-1 - 4 = -5

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

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

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

So 4*n* - 5 = 31.

Move -5 to the right side.

Then 4*n* = 36.

Divide both sides by 4.

Then *n* = 9.

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

So *S*_{9} = (9/2)[-1 + 31].

You can also use the first formula,

by using *d* = 4.

Then *S*_{9} = (9/2)[2⋅(-1) + (9 - 1)⋅4].

-1 + 31 = 30

Cancel the denominator 2.

And reduce the numerator 30 to, 30/2, 15.

9⋅15 = 135

So 135 is the answer.

## Example 3: Find *a*_{n} from *S*_{n}

Think of two cases.

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 equations.

Then*S*_{n} - *S*_{n - 1} = *a*_{n}.

So write*a*_{n} = *S*_{n} - *S*_{n - 1}.

This is true

when *n* = 2, 3, 4, ....

*S*_{n} = *n*^{2} + 2*n*

Then*S*_{n - 1} = (*n* - 1)^{2} + 2(*n* - 1).

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

(*n* - 1)^{2} = *n*^{2} - 2*n* + 1

Square of a difference

2(*n* - 1) = 2*n* - 2

Cancel *n*^{2} and -*n*^{2}.

Cancel +2*n* and -2*n*.

(dark gray terms)

Then (right side) = +2*n* - 1 + 2.

-1 + 2 = +1

So (right side) = 2*n* + 1.

So *a*_{n} = 2*n* + 1

when *n* = 2, 3, 4, ....

Next, think of case 2.

Case 2: *n* = 1

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

So*a*_{1} = 2⋅1 + 1.

2⋅1 + 1

= 2 + 1

= 3

So *a*_{1} = 3.

*S*_{n} = *n*^{2} + 2*n*

So*S*_{1} = 1^{2} + 2⋅1.

So*S*_{1} = 3.

*a*_{1} = 3*S*_{1} = 3

So *a*_{1} = *S*_{1}.

This means*a*_{n} = 2*n* + 1

when *n* = 1.

From case 1,*a*_{n} = 2*n* + 1

when *n* = 2, 3, 4, ....

And from case 2,*a*_{n} = 2*n* + 1

when *n* = 1.

So *a*_{n} = 2*n* + 1

when *n* = 1, 2, 3, 4, ....