Want to learn more about Finite Geometric Series? Take a look at the following step-by-step guide to solve Finite Geometric Series problems.

## Related Topics

- How to Solve Infinite Geometric Series
- How to Solve Geometric Sequences
- How to Solve Arithmetic Sequences

## Step by step guide to solve Finite Geometric Series

- The sum of a geometric series is finite when the absolute value of the ratio is less than \(1\).
- Finite Geometric Series formula: \(\color{blue}{S_{n}=\sum_{i=1}^n ar^{i-1}=a_{1}(\frac{1-r^n}{1-r})}\)

### Finite Geometric Series – Example 1:

Evaluate the geometric series described. \( S_{n}=\sum_{n=1}^5 3^{n-1} \)

**Solution:**

Use this formula: \(S_{n}=\sum_{i=1}^n ar^{i-1} = a_{1} (\frac{ 1- r^n}{1-r})→S_{n}=\sum_{n=1}^5 3^{n-1} =(1)(\frac{ 1- 3^5}{1-3})

→(1)(\frac{ 1- 3^5}{1-3})=121\)

### Finite Geometric Series – Example 2:

Evaluate the geometric series described. \( S_{n}=\sum_{n=1}^3 -4^{n-1} \)

**Solution**:

Use this formula: \(S_{n}=\sum_{i=1}^n ar^{i-1} = a_{1} (\frac{ 1- r^n}{1-r})→S_{n}=\sum_{n=1}^3 -4^{n-1} =(-1)(\frac{ 1- 4^3}{1-4})

→(-1)(\frac{ 1- 4^5}{1-4})=-21\)

### Finite Geometric Series – Example 3:

Evaluate the geometric series described. \( S_{n}=\sum_{n=1}^7 2^{n-1} \)

**Solution**:

Use this formula: \(S_{n}=\sum_{i=1}^n ar^{i-1} = a_{1} (\frac{ 1- r^n}{1-r})→S_{n}=\sum_{n=1}^7 2^{n-1} =1(\frac{ 1- 2^7}{1-2})

→(\frac{ 1- 128}{1-2})=(\frac{-127}{-1})=127\)

### Finite Geometric Series – Example 4:

Evaluate the geometric series described. \( S_{n}=\sum_{n=1}^4 -5^{n-1} \)

**Solution**:

Use this formula: \(S_{n}=\sum_{i=1}^n ar^{i-1} = a_{1} (\frac{ 1- r^n}{1-r})→S_{n}=\sum_{n=1}^4 -5^{n-1} =(-1)(\frac{ 1- 5^4}{1-5})= (-1)(\frac{ 1- 625}{1-5})= (-1)(\frac{ – 624}{-4})= (-1)(\frac{ 624}{4})=-156\)

## Exercises for Solving Finite Geometric Series

### Evaluate each geometric series described.

- \(\color{blue}{1 – 5 + 25 – 125 …, n = 7}\)
- \(\color{blue}{–3 –6 –12 – 24 …, n = 9}\)
- \(\color{blue}{ \sum_{n=1}^8 2, (-2)^{n-1}} \\\ \)
- \(\color{blue}{ \sum_{n=1}^9 4, 3^{n-1} } \\\ \)
- \(\color{blue}{ \sum_{n=1}^{10} 4, (-3)^{n-1} } \\\ \)
- \(\color{blue}{ \sum_{m=1}^9 -2^{m-1} } \\\ \)

- \(\color{blue}{13021}\)
- \(\color{blue}{–1533}\)
- \(\color{blue}{–170}\)
- \(\color{blue}{39364}\)
- \(\color{blue}{–59048}\)
- \(\color{blue}{–511}\)