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 = a₁ + a₂ + a₃ + ... + a_n.

For an arithmetic series S_n,
S_n = (n/2)[2a₁ + (n - 1)d].

a₁: First term
d: Common difference
a_n: nth term

If a_n is given,
then you can also use this formula.

S_n = (n/2)[a₁ + a_n]

a_n: nth term

Example 1

a₁ = 3
d = 5
n = 20

Then S₂₀ = (20/2)[3 + (20 - 1)⋅5].

20/2 = 10

20 - 1 = 19

19⋅5 = 95

6 + 95 = 101

10⋅101 = 1010

So S₂₀ = 1010.

Example 2

The first term is -1.
So a₁ = -1.

7 - 3 = 4
So d = 4.

So a_n = -1 + (n - 1)⋅4.

Arithmetic sequences

(n - 1)⋅4 = 4n - 4

-1 - 4 = -5

So a_n = 4n - 5.

a_n = 4n - 5

And the nth term, a_n, is the last term: 31.

So 4n - 5 = 31.

Move -5 to the right side.

Then 4n = 36.

Divide both sides by 4.

Then n = 9.

a₁ = -1
a_n = 31
n = 9

So S₉ = (9/2)[-1 + 31].

You can also use the first formula,
by using d = 4.

Then S₉ = (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 = a₁ + a₂ + ... + a_{n - 1} + a_n
S_{n - 1} = a₁ + a₂ + ... + 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 = n² + 2n

Then
S_{n - 1} = (n - 1)² + 2(n - 1).

So
a_n = n² + 2n - [(n - 1)² + 2(n - 1)].

(n - 1)² = n² - 2n + 1

Square of a difference

2(n - 1) = 2n - 2

Cancel n² and -n².
Cancel +2n and -2n.
(dark gray terms)

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

-1 + 2 = +1

So (right side) = 2n + 1.

So a_n = 2n + 1
when n = 2, 3, 4, ....

Next, think of case 2.

Case 2: n = 1

See if a₁ = S₁.

a_n = 2n + 1

So
a₁ = 2⋅1 + 1.

2⋅1 + 1
= 2 + 1
= 3

So a₁ = 3.

S_n = n² + 2n

So
S₁ = 1² + 2⋅1.

So
S₁ = 3.

a₁ = 3
S₁ = 3

So a₁ = S₁.

This means
a_n = 2n + 1
when n = 1.

From case 1,
a_n = 2n + 1
when n = 2, 3, 4, ....

And from case 2,
a_n = 2n + 1
when n = 1.

So a_n = 2n + 1
when n = 1, 2, 3, 4, ....

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