# 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:

Sn = (n/2)[2a1 + (n - 1)d]
= (n/2)[a1 + an]

Sn: the sum of '1st term ~ nth term'
a1: first term
d: common difference
an: nth term

## Proof

Sn = a1 + a2 + a3 + ... + an
= a1 + [a1 + 1d] + [a1 + 2d] + ... + [a1 + (n - 1)d]

Arithmetic sequences

Write the same Sn in the opposite order.

Sn = an + an - 1 + an - 2 + ... + a1
= [a1 + (n - 1)d] + [a1 + (n - 2)d] + [a1 + (n - 3)d] + ... + a1

Then (left side) = 2Sn.

In the right side, each term is [2a1 + (n - 1)d].
And there are n terms.

So (right side) = n⋅[2a1 + (n - 1)d].

Divide both sides by 2.

Then Sn = (n/2)[2a1 + (n - 1)d].

2a1 + (n - 1)d = a1 + a1 + (n - 1)d
= a1 + an

So Sn = (n/2)⋅[a1 + an].

## Example 1

a1 = 3, d = 5, d = 20

Use the first formula.

S20 = (20/2)[3 + (20 - 1)⋅5]

## Example 2

a1 = -1, d = +3

So an = -1 + (n - 1)⋅4.

Arithmetic sequences

an = 4n - 5

And the last term, an, is 31.

So 4n - 5 = 31.

n = 9 (number of the terms)

n = 9, a1 = -1, an = 31

Use the second formula.

S9 = (9/2)[-1 + 31]

## Example 3

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

Sn = a1 + a2 + ... + an - 1 + an
Sn - 1 = a1 + a2 + ... + an - 1

Subtract these two.
Sn - Sn - 1 = an

So use an = Sn - Sn - 1
to find an.

an = 2n + 1
(This is true when n = 2, 3, 4, ... .)

Case 2: n = 1

See if an = 2n + 1 is true when n = 1
by checking a1 = S1.

a1 = S1 = 3

So an = 2n + 1 is true when n = 1.

So an = 2n + 1 for all n.
(n = 1, 2, 3, 4, ...)