Geometric Sequence Worksheets - Free Printable

Problem Analysis:

The problem involves a geometric sequence where the number of beans doubles each day. We are tasked with solving three parts related to this scenario:

1. Part (a): Determine how many beans were in the pot on Day 2.
2. Part (b): Calculate how many days it would take for the pot to be full if it was half-full on Day 29.
3. Part (c): Find out how many beans were in the pot on Day 5.

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Solution:

#### Part (a): How many beans were in the pot on Day 2?

- Given: The pot is full on Day 30, and the number of beans doubles each day.
- Key Insight: Since the number of beans doubles every day, the pot must have been half-full on the previous day (Day 29). Similarly, on Day 28, the pot would have been one-fourth full, and so on.

To find the number of beans on Day 2:
- Let the total capacity of the pot be \( C \) (the number of beans when the pot is full).
- On Day 30, the pot is full, so it contains \( C \) beans.
- On Day 29, the pot is half-full, so it contains \( \frac{C}{2} \) beans.
- On Day 28, the pot is one-fourth full, so it contains \( \frac{C}{4} \) beans.
- Continuing this pattern, on Day 2, the pot contains \( \frac{C}{2^{(30-2)}} = \frac{C}{2^{28}} \) beans.

Thus, the number of beans on Day 2 is:
\[
\boxed{\frac{C}{2^{28}}}
\]

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#### Part (b): How many days would it take for the pot to be full if it was half-full on Day 29?

- Given: The pot is half-full on Day 29.
- Key Insight: Since the number of beans doubles each day, if the pot is half-full on Day 29, it will be full on the next day (Day 30).

Thus, it would take:
\[
\boxed{1 \text{ day}}
\]

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#### Part (c): How many beans were in the pot on Day 5?

- Given: The pot is full on Day 30, and the number of beans doubles each day.
- Key Insight: Using the same logic as in Part (a), the number of beans on any given day can be expressed as a fraction of the total capacity \( C \).

To find the number of beans on Day 5:
- On Day 30, the pot is full, so it contains \( C \) beans.
- On Day 29, the pot is half-full, so it contains \( \frac{C}{2} \) beans.
- On Day 28, the pot is one-fourth full, so it contains \( \frac{C}{4} \) beans.
- Continuing this pattern, on Day 5, the pot contains \( \frac{C}{2^{(30-5)}} = \frac{C}{2^{25}} \) beans.

Thus, the number of beans on Day 5 is:
\[
\boxed{\frac{C}{2^{25}}}
\]

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Final Answers:

1. Part (a): \(\boxed{\frac{C}{2^{28}}}\)
2. Part (b): \(\boxed{1 \text{ day}}\)
3. Part (c): \(\boxed{\frac{C}{2^{25}}}\)