Geometric Sequences in the Real World - - Free Printable

Problem Analysis:

The worksheet focuses on geometric sequences, where each term is obtained by multiplying the previous term by a constant ratio \( r \). The tasks involve determining the common ratio, finding subsequent terms, and applying the formula for the \( n \)-th term of a geometric sequence:
\[ a_n = a_1 \cdot r^{n-1} \]

Let's solve each section step by step.

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Section 1: Determine the Common Ratio and Next Term

#### Sequence 1: \( 1, 3, 9, 27, \ldots \)
1. Find the common ratio \( r \):
\[
r = \frac{3}{1} = 3
\]
2. Find the next term:
\[
\text{Next term} = 27 \cdot 3 = 81
\]

#### Sequence 2: \( 40, 20, 10, \ldots \)
1. Find the common ratio \( r \):
\[
r = \frac{20}{40} = \frac{1}{2}
\]
2. Find the next term:
\[
\text{Next term} = 10 \cdot \frac{1}{2} = 5
\]

#### Sequence 3: \( 3, 6, 12, 24, \ldots \)
1. Find the common ratio \( r \):
\[
r = \frac{6}{3} = 2
\]
2. Find the next term:
\[
\text{Next term} = 24 \cdot 2 = 48
\]

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Section 2: Geometric Sequences: Finding the Next Terms

#### Sequence 1: \( 324, 108, 36, \ldots \)
1. Find the common ratio \( r \):
\[
r = \frac{108}{324} = \frac{1}{3}
\]
2. Find the next 2 terms:
\[
\text{Next term} = 36 \cdot \frac{1}{3} = 12
\]
\[
\text{Next term after that} = 12 \cdot \frac{1}{3} = 4
\]

#### Sequence 2: \( -3, -15, -75, \ldots \)
1. Find the common ratio \( r \):
\[
r = \frac{-15}{-3} = 5
\]
2. Find the next 3 terms:
\[
\text{Next term} = -75 \cdot 5 = -375
\]
\[
\text{Next term after that} = -375 \cdot 5 = -1875
\]
\[
\text{Next term after that} = -1875 \cdot 5 = -9375
\]

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Section 3: Geometric Sequences: Finding the \( n \)-th Term

#### Problem 1: Height of a bouncing ball
The sequence is given as:
\[
\begin{array}{c|c}
\text{\# of Bounces} & \text{Height} \\
\hline
1 & 3 \\
2 & 1.8 \\
3 & 1.08 \\
\end{array}
\]

1. Find the common ratio \( r \):
\[
r = \frac{1.8}{3} = 0.6
\]

2. Identify the variables for the formula \( a_n = a_1 \cdot r^{n-1} \):
- \( a_1 = 3 \) (initial height)
- \( r = 0.6 \)
- \( n = 9 \) (9th bounce)

3. Find the height at the 9th bounce (\( a_9 \)):
\[
a_9 = a_1 \cdot r^{9-1} = 3 \cdot (0.6)^8
\]
Calculate \( (0.6)^8 \):
\[
(0.6)^8 = 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \approx 0.0168
\]
\[
a_9 = 3 \cdot 0.0168 \approx 0.0504
\]

#### Problem 2: Growth of bacteria
The sequence is given as:
\[
\begin{array}{c|c}
\text{Hour(s)} & \text{Bacteria} \\
\hline
1 & 250 \\
2 & 500 \\
3 & 1000 \\
\end{array}
\]

1. Find the common ratio \( r \):
\[
r = \frac{500}{250} = 2
\]

2. Identify the variables for the formula \( a_n = a_1 \cdot r^{n-1} \):
- \( a_1 = 250 \) (initial number of bacteria)
- \( r = 2 \)
- \( n = 7 \) (7 hours)

3. Find the number of bacteria after 7 hours (\( a_7 \)):
\[
a_7 = a_1 \cdot r^{7-1} = 250 \cdot 2^6
\]
Calculate \( 2^6 \):
\[
2^6 = 64
\]
\[
a_7 = 250 \cdot 64 = 16000
\]

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Final Answers:

1. Sequence 1: \( r = 3 \), Next term = \( 81 \)
2. Sequence 2: \( r = \frac{1}{2} \), Next term = \( 5 \)
3. Sequence 3: \( r = 2 \), Next term = \( 48 \)
4. Sequence 1 (next 2 terms): \( 12 \) and \( 4 \)
5. Sequence 2 (next 3 terms): \( -375 \), \( -1875 \), and \( -9375 \)
6. Height of bouncing ball at 9th bounce: \( \boxed{0.0504} \)
7. Number of bacteria after 7 hours: \( \boxed{16000} \)

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Thus, the final boxed answers are:
\[
\boxed{0.0504, 16000}
\]