# Geometric Series

How to find the value of a geometric series (sum of a geometric 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 geometric sequence,
Sn = [a(rn - 1)]/[r - 1].

Sn: The sum from a1 to an
a: First term, a1
r: Common ratio

Use this formula when |r| > 1.

(When r = 1, Sn = an.)

Sn = [a(1 - rn)]/[1 - r].

Use this formula when |r| < 1.

## Example 1

### Solution

a = 3
r = 2
n = 7

Then S7 = [3(27 - 1)]/[2 - 1].

27 = 128

Power

2 - 1 = 1

128 - 1 = 127

3⋅127 = 381

## Example 2

### Solution

a1 = 4
So a = 4.

For a geometric sequence,
a3 = a⋅r2.

So
a3 = 4⋅r2.

a3 = 36

So
a3 = 4r2 = 36.

Solve 4r2 = 36.

Divide both sides by 4.

Then r2 = 9.

9 = 32

Square root both sides.

Then r = ±3.

See the two cases.

Case 1: r = 3

a = 4
r = 3

Then S5 = [4(35 - 1)]/[3 - 1].

35 = 243

3 - 1 = 2

Cancel the denominator 2
and reduce the 4 in the numerator to, 4/2, 2.

243 - 1 = 242

2⋅242 = 484

So S5 = 484
for case 1.

See the other case.

Case 2: r = -3

a = 4
r = -3

Then S5 = [4((-3)5 - 1)]/[-3 - 1].

35 = 243
So (-3)5 = -243.

-3 - 1 = -4

4/(-4) = -1

-243 - 1 = -244

-(-244) = 244

So S5 = 244
for case 2.

Case 1: r = 3
S5 = 484

Case 2: r = -3
S5 = 244

So S5 = 484 or 244.

So
484 or 244

## Example 3

### Solution

See the sigma notation.
n goes from 1.

So the first term, a1, is
when n = 1.

So a1 = 5⋅(2/3)1.

5⋅(2/3)1
= 5⋅(2/3)
= 10/3

So a1 = 10/3.

See the term in the sigma:
5⋅(2/3)n.

As n increases,
(2/3) is multiplied:
5⋅(2/3)1, 5⋅(2/3)2, 5⋅(2/3)3, ... .

So the given summation is a geometric series.

So r = 2/3.

n goes from 1 to 4.

So there are 4 terms.

So n = 4.
(This n is for
finding the geometric series S4.)

a = a1 = 10/3
r = 2/3
n = 4

Then S4 = [(10/3)[1 - (2/3)4]]/[1 - 2/3].

|r| < 1
So it's good to use the second formula.

(2/3)4
= 24/34
= 16/81

Power of a Quotient

1 = 3/3

1 = 81/81

3/3 - 2/3 = 1/3

Multiply 3
to both of the numerator and the denominator.

[10/3]⋅3 = 10

[1/3]⋅3 = 1

81/81 - 16/81 = 65/81

10⋅[65/81] = 650/81