Geometric Series worksheet - Free Printable

Problem: Geometric Series Worksheet

We will solve the problems step by step, explaining each part clearly.

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#### Part A: Complete the table for the given geometric series

The general formula for the sum of a geometric series is:
- For a finite geometric series:
\( S_n = a \frac{1 - r^n}{1 - r} \) (if \( |r| < 1 \))
- For an infinite geometric series:
\( S = \frac{a}{1 - r} \) (if \( |r| < 1 \))

Let's analyze each series:

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##### 1) \( 4 + \frac{4}{5} + \frac{4}{25} + \frac{4}{125} + \cdots \)

- Finite or Infinite?: This is an infinite series because it continues indefinitely.
- Common ratio (\( r \)):
The common ratio is found by dividing any term by the previous term:
\( r = \frac{\frac{4}{5}}{4} = \frac{1}{5} \)
- Sum: Since it is an infinite series with \( |r| < 1 \), we use the formula:
\( S = \frac{a}{1 - r} \)
Here, \( a = 4 \) and \( r = \frac{1}{5} \):
\[
S = \frac{4}{1 - \frac{1}{5}} = \frac{4}{\frac{4}{5}} = 4 \cdot \frac{5}{4} = 5
\]

Answer:
- Finite or Infinite?: Infinite
- Common ratio (\( r \)): \( \frac{1}{5} \)
- Sum: \( 5 \)

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##### 2) \( 1 - 3 + 9 - 27 + \cdots \)

- Finite or Infinite?: This is an infinite series because it continues indefinitely.
- Common ratio (\( r \)):
The common ratio is found by dividing any term by the previous term:
\( r = \frac{-3}{1} = -3 \)
- Sum: Since \( |r| = 3 > 1 \), the series does not converge, and the sum is undefined. We denote this as "INF".

Answer:
- Finite or Infinite?: Infinite
- Common ratio (\( r \)): \( -3 \)
- Sum: INF

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##### 3) \( 2 + 1 + \frac{1}{2} + \frac{1}{4} + \cdots \)

- Finite or Infinite?: This is an infinite series because it continues indefinitely.
- Common ratio (\( r \)):
The common ratio is found by dividing any term by the previous term:
\( r = \frac{1}{2} \)
- Sum: Since it is an infinite series with \( |r| < 1 \), we use the formula:
\( S = \frac{a}{1 - r} \)
Here, \( a = 2 \) and \( r = \frac{1}{2} \):
\[
S = \frac{2}{1 - \frac{1}{2}} = \frac{2}{\frac{1}{2}} = 2 \cdot 2 = 4
\]

Answer:
- Finite or Infinite?: Infinite
- Common ratio (\( r \)): \( \frac{1}{2} \)
- Sum: \( 4 \)

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##### 4) \( -3 - 6 - 12 - 24, \ldots, n = 8 \)

- Finite or Infinite?: This is a finite series because it has a specific number of terms (\( n = 8 \)).
- Common ratio (\( r \)):
The common ratio is found by dividing any term by the previous term:
\( r = \frac{-6}{-3} = 2 \)
- Sum: Since it is a finite series, we use the formula:
\( S_n = a \frac{1 - r^n}{1 - r} \)
Here, \( a = -3 \), \( r = 2 \), and \( n = 8 \):
\[
S_8 = -3 \frac{1 - 2^8}{1 - 2} = -3 \frac{1 - 256}{-1} = -3 \frac{-255}{-1} = -3 \cdot 255 = -765
\]

Answer:
- Finite or Infinite?: Finite
- Common ratio (\( r \)): \( 2 \)
- Sum: \( -765 \)

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##### 5) \( 4 + 20 + 100 + \cdots + 312500 \)

- Finite or Infinite?: This is a finite series because it has a specific last term.
- Common ratio (\( r \)):
The common ratio is found by dividing any term by the previous term:
\( r = \frac{20}{4} = 5 \)
- Sum: To find the sum, we first need to determine the number of terms (\( n \)). The general term of a geometric series is given by:
\( a_n = a \cdot r^{n-1} \)
Here, \( a = 4 \), \( r = 5 \), and \( a_n = 312500 \):
\[
312500 = 4 \cdot 5^{n-1} \implies 5^{n-1} = \frac{312500}{4} = 78125
\]
Since \( 78125 = 5^7 \), we have:
\( 5^{n-1} = 5^7 \implies n-1 = 7 \implies n = 8 \)
Now, we use the finite sum formula:
\( S_n = a \frac{1 - r^n}{1 - r} \)
Here, \( a = 4 \), \( r = 5 \), and \( n = 8 \):
\[
S_8 = 4 \frac{1 - 5^8}{1 - 5} = 4 \frac{1 - 390625}{-4} = 4 \cdot \frac{-390624}{-4} = 4 \cdot 97656 = 390624
\]

Answer:
- Finite or Infinite?: Finite
- Common ratio (\( r \)): \( 5 \)
- Sum: \( 390624 \)

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#### Part B: Determine the needed information to find the sum of the geometric series

##### 1) Find the sum of the first 8 terms of the geometric series: \( -1 + 2 - 4 + 8 \cdots \)

- First term (\( a_1 \)): \( -1 \)
- Number of terms (\( n \)): \( 8 \)
- Common ratio (\( r \)):
\( r = \frac{2}{-1} = -2 \)
- Sum: Use the finite sum formula:
\( S_n = a \frac{1 - r^n}{1 - r} \)
Here, \( a = -1 \), \( r = -2 \), and \( n = 8 \):
\[
S_8 = -1 \frac{1 - (-2)^8}{1 - (-2)} = -1 \frac{1 - 256}{1 + 2} = -1 \frac{-255}{3} = -1 \cdot (-85) = 85
\]

Answer:
- \( a_1 = -1 \)
- \( n = 8 \)
- \( r = -2 \)
- Final Answer: The sum is \( 85 \)

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##### 2) Find the sum of the geometric series: \( 4 - \frac{4}{3} + \frac{4}{9} - \frac{4}{27} \cdots \)

- First term (\( a_1 \)): \( 4 \)
- Number of terms (\( n \)): This is an infinite series because it continues indefinitely.
- Common ratio (\( r \)):
\( r = \frac{-\frac{4}{3}}{4} = -\frac{1}{3} \)
- Sum: Since it is an infinite series with \( |r| < 1 \), we use the formula:
\( S = \frac{a}{1 - r} \)
Here, \( a = 4 \) and \( r = -\frac{1}{3} \):
\[
S = \frac{4}{1 - \left(-\frac{1}{3}\right)} = \frac{4}{1 + \frac{1}{3}} = \frac{4}{\frac{4}{3}} = 4 \cdot \frac{3}{4} = 3
\]

Answer:
- \( a_1 = 4 \)
- \( n = \text{INF} \)
- \( r = -\frac{1}{3} \)
- Final Answer: The sum is \( 3 \)

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Final Answers:

#### Part A:
1. Infinite, \( \frac{1}{5} \), \( 5 \)
2. Infinite, \( -3 \), INF
3. Infinite, \( \frac{1}{2} \), \( 4 \)
4. Finite, \( 2 \), \( -765 \)
5. Finite, \( 5 \), \( 390624 \)

#### Part B:
1. \( a_1 = -1 \), \( n = 8 \), \( r = -2 \), Final Answer: \( 85 \)
2. \( a_1 = 4 \), \( n = \text{INF} \), \( r = -\frac{1}{3} \), Final Answer: \( 3 \)

Boxed Final Answer:

\[
\boxed{
\begin{array}{c}
\text{Part A:} \\
1. \text{Infinite, } \frac{1}{5}, 5 \\
2. \text{Infinite, } -3, \text{INF} \\
3. \text{Infinite, } \frac{1}{2}, 4 \\
4. \text{Finite, } 2, -765 \\
5. \text{Finite, } 5, 390624 \\
\\
\text{Part B:} \\
1. a_1 = -1, n = 8, r = -2, \text{Final Answer: } 85 \\
2. a_1 = 4, n = \text{INF}, r = -\frac{1}{3}, \text{Final Answer: } 3 \\
\end{array}
}
\]