# 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 *S*_{n},*S*_{n} = *a*_{1}(1 - *r*^{n})/(1 - *r*).*a*_{1}: First term*r*: Common ratio

## Example 1

*a*_{1} = 3*r* = 2*n* = 7

Then *S*_{7} = 3(1 - 2^{7})/(1 - 2).

3(1 - 2^{7})/(1 - 2) = 3(2^{7} - 1)/(2 - 1)

2^{7} = 128

128 - 1 = 127

3⋅127 = 381

So *S*_{7} = 381.

## Example 2

*a*_{1} = 4

Then *a*_{3} = 4⋅*r*^{2} = 36.

Geometric sequences

Solve 4⋅*r*^{2} = 36.

Divide both side by 4.

Then *r*^{2} = 9.

9 = (±3)^{2}

So *r* = ±3.

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

Case 1: *r* = 3*a*_{1} = 4*n* = 5

Then *S*_{5} = 4(1 - 3^{5})/(1 - 3).

4(1 - 3^{5})/(1 - 3) = 4(3^{5} - 1)/(3 - 1)

3^{5} = 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,*S*_{5} = 484.

Case 2: *r* = -3*a*_{1} = 4*n* = 5

Then *S*_{5} = 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,*S*_{5} = 244.

For case 1,*S*_{5} = 484.

For case 2,*S*_{5} = 244.

So *S*_{5} = 484 or 244.

## Example 3

*a*_{n} = 5⋅(2/3)^{n}

So *a*_{1} = 5⋅(2/3).

So *a*_{1} = 10/3.

*a*_{n} = 5⋅(2/3)^{n}

As *n* increases +1,*a*_{n} 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

*a*_{1} = 10/3*r* = 2/3*n* = 4

Then *S*_{4} = (10/3)[1 - (2/3)^{4}]/(1 - 2/3)

(2/3)^{4}

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

= 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.