Geometric Sequence Word Problems worksheet with five math problems.

Problem Analysis and Solutions

The problems provided involve geometric sequences. A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

The general formula for the \( n \)-th term of a geometric sequence is:
\[
a_n = a_1 \cdot r^{n-1}
\]
where:
- \( a_n \) is the \( n \)-th term,
- \( a_1 \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the term number.

Let's solve each problem step by step.

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Problem 1:

A manufacturing unit produced 112, 336, and 1008 Easter baskets on days one, two, and three, respectively. If the unit continues to manufacture baskets at the same rate, how many baskets will it produce on the 4th, 5th, and 6th day?

#### Step 1: Identify the first term and common ratio.
- The first term \( a_1 = 112 \).
- The second term \( a_2 = 336 \).
- The third term \( a_3 = 1008 \).

The common ratio \( r \) is calculated as:
\[
r = \frac{a_2}{a_1} = \frac{336}{112} = 3
\]
Verify with the third term:
\[
a_3 = a_2 \cdot r = 336 \cdot 3 = 1008
\]
The common ratio is consistent.

#### Step 2: Use the formula for the \( n \)-th term.
The formula is:
\[
a_n = a_1 \cdot r^{n-1}
\]

- For the 4th day (\( n = 4 \)):
\[
a_4 = 112 \cdot 3^{4-1} = 112 \cdot 3^3 = 112 \cdot 27 = 3024
\]

- For the 5th day (\( n = 5 \)):
\[
a_5 = 112 \cdot 3^{5-1} = 112 \cdot 3^4 = 112 \cdot 81 = 9072
\]

- For the 6th day (\( n = 6 \)):
\[
a_6 = 112 \cdot 3^{6-1} = 112 \cdot 3^5 = 112 \cdot 243 = 27216
\]

#### Final Answer:
\[
\boxed{3024, 9072, 27216}
\]

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Problem 2:

Joanne and her friends went on a hiking vacation. They covered 2 miles on the first day, 4 miles on the second day, and 8 miles on the third day. The hiking distance continued in a geometric sequence. How many miles did they hike on the 4th day?

#### Step 1: Identify the first term and common ratio.
- The first term \( a_1 = 2 \).
- The second term \( a_2 = 4 \).
- The third term \( a_3 = 8 \).

The common ratio \( r \) is:
\[
r = \frac{a_2}{a_1} = \frac{4}{2} = 2
\]
Verify with the third term:
\[
a_3 = a_2 \cdot r = 4 \cdot 2 = 8
\]
The common ratio is consistent.

#### Step 2: Use the formula for the \( n \)-th term.
The formula is:
\[
a_n = a_1 \cdot r^{n-1}
\]

- For the 4th day (\( n = 4 \)):
\[
a_4 = 2 \cdot 2^{4-1} = 2 \cdot 2^3 = 2 \cdot 8 = 16
\]

#### Final Answer:
\[
\boxed{16}
\]

---

Problem 3:

Nina and Zoe sold 3 cookies on Monday, 9 cookies on Tuesday, and 27 cookies on Wednesday. The sales continued in a geometric sequence. How many cookies did they sell on Friday?

#### Step 1: Identify the first term and common ratio.
- The first term \( a_1 = 3 \).
- The second term \( a_2 = 9 \).
- The third term \( a_3 = 27 \).

The common ratio \( r \) is:
\[
r = \frac{a_2}{a_1} = \frac{9}{3} = 3
\]
Verify with the third term:
\[
a_3 = a_2 \cdot r = 9 \cdot 3 = 27
\]
The common ratio is consistent.

#### Step 2: Use the formula for the \( n \)-th term.
The formula is:
\[
a_n = a_1 \cdot r^{n-1}
\]

- Monday is day 1, Tuesday is day 2, Wednesday is day 3, Thursday is day 4, and Friday is day 5.
- For Friday (\( n = 5 \)):
\[
a_5 = 3 \cdot 3^{5-1} = 3 \cdot 3^4 = 3 \cdot 81 = 243
\]

#### Final Answer:
\[
\boxed{243}
\]

---

Problem 4:

Charles plants 8 seeds on day 1, 16 seeds on day 2, 32 seeds on day 3, and so on in a geometric sequence. How many seeds will he plant on the 10th day?

#### Step 1: Identify the first term and common ratio.
- The first term \( a_1 = 8 \).
- The second term \( a_2 = 16 \).
- The third term \( a_3 = 32 \).

The common ratio \( r \) is:
\[
r = \frac{a_2}{a_1} = \frac{16}{8} = 2
\]
Verify with the third term:
\[
a_3 = a_2 \cdot r = 16 \cdot 2 = 32
\]
The common ratio is consistent.

#### Step 2: Use the formula for the \( n \)-th term.
The formula is:
\[
a_n = a_1 \cdot r^{n-1}
\]

- For the 10th day (\( n = 10 \)):
\[
a_{10} = 8 \cdot 2^{10-1} = 8 \cdot 2^9 = 8 \cdot 512 = 4096
\]

#### Final Answer:
\[
\boxed{4096}
\]

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Problem 5:

Mr. Botham is conducting an experiment to observe the growth of bacteria. He records the original count of the bacteria as 15. If the number increases by 2% every hour, how many bacteria will there be after 3 hours?

#### Step 1: Understand the growth pattern.
The bacteria grow by 2% every hour, which means the population multiplies by \( 1 + 0.02 = 1.02 \) each hour. This forms a geometric sequence where:
- The first term \( a_1 = 15 \).
- The common ratio \( r = 1.02 \).

#### Step 2: Use the formula for the \( n \)-th term.
The formula is:
\[
a_n = a_1 \cdot r^{n-1}
\]

- After 3 hours (\( n = 4 \) because the initial count is at hour 0):
\[
a_4 = 15 \cdot (1.02)^{4-1} = 15 \cdot (1.02)^3
\]

Calculate \( (1.02)^3 \):
\[
(1.02)^3 = 1.02 \cdot 1.02 \cdot 1.02 \approx 1.061208
\]

Now calculate \( a_4 \):
\[
a_4 = 15 \cdot 1.061208 \approx 15.91812
\]

Round to the nearest whole number:
\[
a_4 \approx 16
\]

#### Final Answer:
\[
\boxed{16}
\]

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Final Answers:

1. \(\boxed{3024, 9072, 27216}\)
2. \(\boxed{16}\)
3. \(\boxed{243}\)
4. \(\boxed{4096}\)
5. \(\boxed{16}\)