# Arithmetic Series

How to find the value of an arithmetic series (the sum of an arithmetic sequence): formula, 3 examples, and their solutions.

## Formula

### Formula

A series means

the sum of the terms of a sequence.

S_{n} = a_{1} + a_{2} + a_{3} + ... + a_{n}

For an arithmetic sequence,

S_{n} = [n/2][2a + (n - 1)d].

S_{n}: The sum from a_{1} to a_{n}

a: First term, a_{1}

d: Common difference

If a_{1} and a_{n} are given,

you can also use this formula.

S_{n} = [n/2][a_{1} + a_{n}]

## Example 1

### Example

### Solution

a = 3

d = 5

n = 20

Then S_{20} = [20/2][2⋅3 + 19⋅5].

20/2 = 10

2⋅3 = 6

+19⋅5 = +95

6 + 95 = 101

10⋅101 = 1010

So S_{20} = 1010.

## Example 2

### Example

### Solution

To find the number of the terms, n,

first find a_{n}.

Write -1, 3, 7, 10.

The first term, a, is -1.

-1 + 4 = 3

3 + 4 = 7

7 + 4 = 11

So write, the d, +4

between the terms.

a = -1

d = 4

So a_{n} = -1 + (n - 1)⋅4.

+(n - 1)⋅4 = +4n - 4

Multiply a Monomial and a Polynomial

-1 + 4n - 4 = 4n - 5

So a_{n} = 4n - 5.

The last term, a_{n}, is 31.

a_{n} = 4n - 5

So 4n - 5 = 31.

Move -5 to the right side.

Then 4n = 36.

Divide both sides by 4.

Then n = 9.

a = -1

d = 4

n = 9

Then S_{9} = [9/2][2⋅(-1) + 8⋅4].

You can also use the other formula.

n = 9

a_{1} = -1

a_{n} = 31

Then S_{9} = [9/2][-1 + 31].

2⋅(-1) = -2

+8⋅4 = +32

-2 + 32 = 30

Cancel the denominator 2

and reduce 30 to, 30/2, 15.

9⋅15 = 135

So 135 is the answer.

## Example 3

### Example

### Solution

S_{n} is given.

And it says to find a_{n}.

In this case,

think of two cases:

1) n = 2, 3, 4, ... and 2) n = 1.

Case 1: n = 2, 3, 4, ...

S_{n} = a_{1} + a_{2} + a_{3} + ... + a_{n - 1} + a_{n}

S_{n - 1} = a_{1} + a_{2} + a_{3} + ... + a_{n - 1}

Subtract these two equations.

Then S_{n} - S_{n - 1} = a_{n}.

So set a_{n} = S_{n} - S_{n - 1}.

This is true

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

S_{n} = n^{2} + 2n

So write n^{2} + 2n.

Write -.

S_{n} = n^{2} + 2n

So

S_{n - 1} = (n - 1)^{2} + 2(n - 1).

So

S_{n} - S_{n - 1}

= n^{2} + 2n - [(n - 1)^{2} + 2(n - 1)].

(n - 1)^{2} = n^{2} - 2⋅n⋅1 + 1^{2}

Square of a Difference: (a - b)^{2}

+2(n - 1) = +2n - 2

-2⋅n⋅1 = -2n

+1^{2} = +1

Cancel n^{2}.

Cancel +2n.

+1 - 2 = -1

-[-2n - 1] = 2n + 1

So a_{n} = 2n + 1

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

Next, see case 2: n = 1.

When n = 1,

a_{n} = a_{1}.

S_{1} = a_{1}

So a_{1} = S_{1}.

S_{n} = n^{2} + 2n

So S_{1} = 1^{2} + 2⋅1.

1^{2} = 1

+2⋅1 = +2

1 + 2 = 3

So a_{1} = 3.

You got a_{1} = 3.

Then see if a_{n} = 2n + 1,

the a_{n} from case 1,

is true for n = 1.

In other words,

see if a_{1} from a_{n} = 2n + 1

is equal to 3.

a_{1} = 2⋅1 + 1

2⋅1 = 2

2 + 1 = 3

So a_{1} = 3.

This a_{1} is from a_{n} = 2n + 1.

a_{1} from S_{1}

is 3.

And a_{1} from a_{n} = 2n + 1

is also 3.

a_{1} from a_{n}

is equal to

a_{1} from S_{1}.

Then a_{n} = 2n + 1

is true for n = 1.

(If a_{1} from a_{n}

is not equal to

a_{1} from S_{1},

then a_{1} = S_{1}

and a_{n} is not true for n = 1.)

Case 1:

a_{n} = 2n + 1

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

Case 2:

a_{n} = 2n + 1

(n = 1)

So a_{n} = 2n + 1.

So a_{n} = 2n + 1.