# Geometric Series

How to find the value of a geometric series (sum of a geometric sequence): formula, 3 examples, and their solutions.

## Formula

A series means

the sum of the terms of a sequence.

S_{n} = a_{1} + a_{2} + a_{3} + ... + a_{n}

For an geometric sequence,

S_{n} = [a(r^{n} - 1)]/[r - 1].

S_{n}: The sum from a_{1} to a_{n}

a: First term, a_{1}

r: Common ratio

Use this formula when |r| > 1.

(When r = 1, S_{n} = an.)

S_{n} = [a(1 - r^{n})]/[1 - r].

Use this formula when |r| < 1.

## Examplea = 3, r = 2, n = 7, S_{7} = ?

a = 3

r = 2

n = 7

Then S_{7} = [3(2^{7} - 1)]/[2 - 1].

2^{7} = 128

Power

2 - 1 = 1

128 - 1 = 127

3⋅127 = 381

So 381 is the answer.

## Examplea_{1} = 4, a_{4} = 36, S_{5} = ?

a_{1} = 4

So a = 4.

For a geometric sequence,

a_{3} = a⋅r^{2}.

So

a_{3} = 4⋅r^{2}.

a_{3} = 36

So

a_{3} = 4r^{2} = 36.

Solve 4r^{2} = 36.

Divide both sides by 4.

Then r^{2} = 9.

9 = 3^{2}

Square root both sides.

Then r = ±3.

See the two cases.

Case 1: r = 3

a = 4

r = 3

Then S_{5} = [4(3^{5} - 1)]/[3 - 1].

3^{5} = 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 S_{5} = 484

for case 1.

See the other case.

Case 2: r = -3

a = 4

r = -3

Then S_{5} = [4((-3)^{5} - 1)]/[-3 - 1].

3^{5} = 243

So (-3)^{5} = -243.

-3 - 1 = -4

4/(-4) = -1

-243 - 1 = -244

-(-244) = 244

So S_{5} = 244

for case 2.

Case 1: r = 3

S_{5} = 484

Case 2: r = -3

S_{5} = 244

So S_{5} = 484 or 244.

So

484 or 244

is the answer.

## Example∑_{n = 1}^{4} 5⋅(2/3)^{n}

See the sigma notation.

n goes from 1.

So the first term, a_{1}, is

when n = 1.

So a_{1} = 5⋅(2/3)^{1}.

5⋅(2/3)^{1}

= 5⋅(2/3)

= 10/3

So a_{1} = 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 S_{4}.)

a = a_{1} = 10/3

r = 2/3

n = 4

Then S_{4} = [(10/3)[1 - (2/3)^{4}]]/[1 - 2/3].

|r| < 1

So it's good to use the second formula.

(2/3)^{4}

= 2^{4}/3^{4}

= 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

So 650/81 is the answer.