 # Geometric Series How to find the value of a geometric series: formula, examples, and their solutions.

## Formula A geometric series is
the sum of the terms of the geometric sequence.

Geometric sequences

For a geometric series Sn,
Sn = a1(1 - rn)/(1 - r).

a1: First term
r: Common ratio

## Example 1 a1 = 3
r = 2
n = 7

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

3(1 - 27)/(1 - 2) = 3(27 - 1)/(2 - 1)

27 = 128

128 - 1 = 127

3⋅127 = 381

So S7 = 381.

## Example 2 a1 = 4

Then a3 = 4⋅r2 = 36.

Geometric sequences

Solve 4⋅r2 = 36.

Divide both side by 4.

Then r2 = 9.

9 = (±3)2

So r = ±3.

Then you should think of two cases:
r = 3 and r = -3.

Case 1: r = 3

a1 = 4
n = 5

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

4(1 - 35)/(1 - 3) = 4(35 - 1)/(3 - 1)

35 = 243

3 - 1 = 2

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

243 - 1 = 242

2⋅242 = 484

So, for case 1,
S5 = 484.

Case 2: r = -3

a1 = 4
n = 5

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

(-3)5 = -243

1 - (-3) = 1 + 3

1 - (-243) = 1 + 243

1 + 3 = 4

Cancel the numerator 4
and cancel the denominator 4.

1 + 243 = 244

So, for case 2,
S5 = 244.

For case 1,
S5 = 484.

For case 2,
S5 = 244.

So S5 = 484 or 244.

## Example 3 an = 5⋅(2/3)n

So a1 = 5⋅(2/3).

So a1 = 10/3.

an = 5⋅(2/3)n

As n increases +1,
an changes ×(2/3).

So the given summation shows
a geometric series.

And r = 2/3.

Geometric sequences

See the given summation.
n goes from 1 to 4.

So there are 4 terms.

So n = 4.

Sigma notation

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

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

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

1 - 2/3 = 1/3

Multiply 3
to both of the numerator and the denominator.

Then (right side) = 10[1 - 16/81].

1 = 81/81

81/81 - 16/81 = 65/81

10⋅65 = 650

So (given) = 650/81.