Arithmetic Series

Arithmetic Series

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

Formulas

An arithmetic series is the sum of the terms of the arithmetic sequence. Sn = (n/2)[2*a1 + (n - 1)d] = (n/2)[a1 + an]. a1: First term, d: Common difference, an: nth term

A series means the sum of the sequence.

So an arithmetic series is
the sum of the terms of the arithmetic sequence.

Arithmetic sequences

Sn is the symbol of a series:
Sn = a1 + a2 + a3 + ... + an.

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

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

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

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

an: nth term

Example 1

For the given arithmetic sequence, find Sn. a1 = 3. d = 5, n = 20

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

Find the value of the given arithmetic series. -1 + 3 + 7 + 10 + ... + 31

The first term is -1.
So a1 = -1.

7 - 3 = 4
So d = 4.

So an = -1 + (n - 1)⋅4.

Arithmetic sequences

(n - 1)⋅4 = 4n - 4

-1 - 4 = -5

So an = 4n - 5.

an = 4n - 5

And the nth term, an, 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
an = 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 an from Sn

For the given arithmetic series, find an. Sn = n^2 + 2n

Think of two cases.

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

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

Subtract these two equations.
Then
Sn - Sn - 1 = an.

So write
an = Sn - Sn - 1.

This is true
when n = 2, 3, 4, ....

Sn = n2 + 2n

Then
Sn - 1 = (n - 1)2 + 2(n - 1).

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

Next, think of case 2.

Case 2: n = 1

See if a1 = S1.

an = 2n + 1

So
a1 = 2⋅1 + 1.

2⋅1 + 1
= 2 + 1
= 3

So a1 = 3.

Sn = n2 + 2n

So
S1 = 12 + 2⋅1.

So
S1 = 3.

a1 = 3
S1 = 3

So a1 = S1.

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

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

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

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