## How to Solve Finite Geometric Series

### Finite Geometric Series

There are a finite set of numbers in a finite geometric series. This means there will be a first and last term for the series. Finite geometric series are convergent.

### Finite Geometric Formula

Use the formula to find the sum of a finite geometric series. $$S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1}$$, when $$r \ ≠ \ 1$$

Where $$a$$ is the first term, $$n$$ is the number of terms, and $$r$$ is the common ratio.

### Example

Find the total of the first $$6$$ terms of the geometric series if $$a \ = \ 5$$ and $$r \ = \ 3$$.

$$S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ ⇒ \ \frac{5(3^6 \ - \ 1)}{3 \ - \ 1} \ = \ 1820$$

### Exercises for Finite Geometric Series

1) Evaluate the geometric series described: $$-5, \ 15, \ -45, \ 135, \ -405, \ ..., \ n \ = \ 7$$

2) Evaluate the geometric series described: $$9, \ 18, \ , \ 36, \ 72, \ ..., \ n \ = \ 9$$

3) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 9} \ 7(-2)^{n \ - \ 1}$$

4) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 5} \ 5(3)^{n \ - \ 1}$$

5) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 6} \ 3(-\frac{1}{2})^{n \ - \ 1}$$

6) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 4} \ 8(\frac{1}{3})^{n \ - \ 1}$$

7) Evaluate the geometric series described: $$1, \ 4, \ 16, \ 64, \ 256, \ ..., \ n \ = \ 6$$

8) Evaluate the geometric series described: $$1024, \ 256, \ 64, \ 16, \ 4, \ ..., \ n \ = \ 6$$

9) Evaluate the geometric series described: $$7, \ 14, \ 28, \ 56, \ 112, \ ..., \ n \ = \ 4$$

10) Evaluate the geometric series described: $$54, \ 18, \ 6, \ 2, \ \frac{2}{3}, \ ..., \ n \ = \ 4$$

1) Evaluate the geometric series described: $$-5, \ 15, \ -45, \ 135, \ -405, \ ..., \ n \ = \ 7$$

$$\color{red}{r \ = \ \frac{15}{-5} \ = \ -3}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{-5((-3)^7 \ - \ 1)}{-3 \ - \ 1} \ = \ \frac{-5(-2187 \ - \ 1)}{-3 \ - \ 1} \ = \ 2735}$$

2) Evaluate the geometric series described: $$9, \ 18, \ , \ 36, \ 72, \ ..., \ n \ = \ 9$$

$$\color{red}{r \ = \ \frac{18}{9} \ = \ 2}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{9(2^9 \ - \ 1)}{2 \ - \ 1} \ = \ 9(512 \ - \ 1) \ = \ 4599}$$

3) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 9} \ 7(-2)^{n \ - \ 1}$$

$$\color{red}{\sum_{n \ = \ 1}^{n \ = \ 9} \ 7(-2)^{n \ - \ 1} \ ⇒ \ r \ = \ -2, \ a \ = \ 7}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{7((-2)^9 \ - \ 1)}{-2 \ - \ 1} \ = \ \frac{7(-512 \ - \ 1)}{-3} \ = \ 1197}$$

4) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 5} \ 5(3)^{n \ - \ 1}$$

$$\color{red}{\sum_{n \ = \ 1}^{n \ = \ 5} \ 5(3)^{n \ - \ 1} \ ⇒ \ r \ = \ 3, \ a \ = \ 5}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{5(3^5 \ - \ 1)}{3 \ - \ 1} \ = \ \frac{5(243 \ - \ 1)}{2} \ = \ 605}$$

5) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 6} \ 3(-\frac{1}{2})^{n \ - \ 1}$$

$$\color{red}{\sum_{n \ = \ 1}^{n \ = \ 6} \ 3(-\frac{1}{2})^{n \ - \ 1} \ ⇒ \ r \ = \ -\frac{1}{2} \ , \ a \ = \ 3}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{3((-\frac{1}{2})^5 \ - \ 1)}{-\frac{1}{2} \ - \ 1} \ = \ \frac{3(-\frac{1}{32} \ - \ 1)}{-\frac{3}{2}} \ = \ \frac{66}{32} \ = \ \frac{33}{16}}$$

6) Evaluate the geometric series described: $$\sum_{n \ = \ 1}^{n \ = \ 4} \ 8(\frac{1}{3})^{n \ - \ 1}$$

$$\color{red}{\sum_{n \ = \ 1}^{n \ = \ 4} \ 8(\frac{1}{3})^{n \ - \ 1} \ ⇒ \ r \ = \ \frac{1}{3} \ , \ a \ = \ 8}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{8((\frac{1}{3})^4 \ - \ 1)}{\frac{1}{3} \ - \ 1} \ = \ \frac{8(\frac{1}{81} \ - \ 1)}{-\frac{2}{3}} \ = \ \frac{320}{27}}$$

7) Evaluate the geometric series described: $$1, \ 4, \ 16, \ 64, \ 256, \ ..., \ n \ = \ 6$$

$$\color{red}{r \ = \ \frac{4}{1} \ = \ 4}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{1((4)^6 \ - \ 1)}{4 \ - \ 1} \ = \ \frac{4096 \ - \ 1}{3} \ = \ 1365}$$

8) Evaluate the geometric series described: $$1024, \ 256, \ 64, \ 16, \ 4, \ ..., \ n \ = \ 6$$

$$\color{red}{r \ = \ \frac{256}{1024} \ = \ \frac{1}{4}}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{1024((\frac{1}{4} \ )^6 \ - \ 1)}{\frac{1}{4} \ - \ 1} \ = \ \frac{1024(\frac{1}{4096} \ - \ 1}{-\frac{3}{4}} \ = \ 1365}$$

9) Evaluate the geometric series described: $$7, \ 14, \ 28, \ 56, \ 112, \ ..., \ n \ = \ 4$$

$$\color{red}{r \ = \ \frac{14}{7} \ = \ 2}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{7((2)^4 \ - \ 1)}{2 \ - \ 1} \ = \ 7(16 \ - \ 1) \ = \ 105}$$

10) Evaluate the geometric series described: $$54, \ 18, \ 6, \ 2, \ \frac{2}{3}, \ ..., \ n \ = \ 4$$

$$\color{red}{r \ = \ \frac{18}{54} \ = \ \frac{1}{3}}$$
$$\color{red}{S_n \ = \ \frac{a(r^n \ - \ 1)}{r \ - \ 1} \ = \ \frac{54((\frac{1}{3})^4 \ - \ 1)}{\frac{1}{3} \ - \ 1} \ = \ 80}$$

## Finite Geometric Series Practice Quiz

### SAT Math in 30 Days

$22.99$14.99

### 5 CLEP College Algebra Practice Tests

$20.99$15.99

### SSAT Upper Level Math Study Guide

$20.99$15.99

### 7th Grade GMAS Math Workbook

$18.99$12.99