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.

Sn = a1 + a2 + a3 + ... + an

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

Sn: The sum from a1 to an
a: First term, a1
d: Common difference

If a1 and an are given,
you can also use this formula.

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

Example 1

Example

Solution

a = 3
d = 5
n = 20

Then S20 = [20/2][2⋅3 + 19⋅5].

20/2 = 10

2⋅3 = 6
+19⋅5 = +95

6 + 95 = 101

10⋅101 = 1010

So S20 = 1010.

Example 2

Example

Solution

To find the number of the terms, n,
first find an.

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 an = -1 + (n - 1)⋅4.

+(n - 1)⋅4 = +4n - 4

Multiply a Monomial and a Polynomial

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

So an = 4n - 5.

The last term, an, is 31.

an = 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 S9 = [9/2][2⋅(-1) + 8⋅4].

You can also use the other formula.

n = 9
a1 = -1
an = 31

Then S9 = [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

Sn is given.
And it says to find an.

In this case,
think of two cases:
1) n = 2, 3, 4, ... and 2) n = 1.

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

Sn = a1 + a2 + a3 + ... + an - 1 + an
Sn - 1 = a1 + a2 + a3 + ... + an - 1

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

So set an = Sn - Sn - 1.

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

Sn = n2 + 2n

So write n2 + 2n.

Write -.

Sn = n2 + 2n

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

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

(n - 1)2 = n2 - 2⋅n⋅1 + 12

Square of a Difference: (a - b)2

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

-2⋅n⋅1 = -2n
+12 = +1

Cancel n2.
Cancel +2n.

+1 - 2 = -1

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

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

Next, see case 2: n = 1.

When n = 1,
an = a1.

S1 = a1

So a1 = S1.

Sn = n2 + 2n

So S1 = 12 + 2⋅1.

12 = 1
+2⋅1 = +2

1 + 2 = 3

So a1 = 3.

You got a1 = 3.

Then see if an = 2n + 1,
the an from case 1,
is true for n = 1.

In other words,
see if a1 from an = 2n + 1
is equal to 3.

a1 = 2⋅1 + 1

2⋅1 = 2

2 + 1 = 3

So a1 = 3.

This a1 is from an = 2n + 1.

a1 from S1
is 3.

And a1 from an = 2n + 1
is also 3.

a1 from an
is equal to
a1 from S1.

Then an = 2n + 1
is true for n = 1.

(If a1 from an
is not equal to
a1 from S1,

then a1 = S1
and an is not true for n = 1.)

Case 1:
an = 2n + 1
(n = 2, 3, 4, ...)

Case 2:
an = 2n + 1
(n = 1)

So an = 2n + 1.

So an = 2n + 1.