? Geometric Sequences Activity | KS3 Algebra | Beyond Maths - Free Printable

Problem Overview:

The image contains two types of puzzles related to geometric sequences:

1. Grid Puzzle: Fill in the blank boxes in a grid such that consecutive unshaded boxes in any row or column form a geometric sequence.
2. Cross-Number Puzzle: Use the first few terms of given geometric sequences to find specific terms and fill them into a crossword-like grid.

Below, I will explain how to solve each type of puzzle step by step.

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Part 1: Grid Puzzle

#### Objective:
Fill in the blank boxes in the grid so that consecutive unshaded boxes in any row or column form a geometric sequence.

#### Key Concepts:
- 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.
- If three consecutive terms in a geometric sequence are \(a\), \(b\), and \(c\), then \(b^2 = ac\) (since \(b = ar\) and \(c = ar^2\)).

#### Solution Approach:
1. Identify the given numbers in the grid.
2. Use the property of geometric sequences (\(b^2 = ac\)) to determine missing terms.
3. Ensure that all rows and columns satisfy the geometric sequence condition.

#### Step-by-Step Solution:
Let's analyze the grid row by row and column by column.

##### Row 1:
- Given: \(3\), \(4.8\), \(\_\), \(96\)
- Let the missing term be \(x\).
- Using the geometric sequence property: \(4.8^2 = 3 \cdot x\).
\[
4.8^2 = 23.04 \quad \text{and} \quad 3 \cdot x = 23.04 \implies x = \frac{23.04}{3} = 7.68
\]
- So, the row becomes: \(3\), \(4.8\), \(7.68\), \(96\).

##### Row 2:
- Given: \(\_\), \(\_\), \(72\), \(\_\), \(75\)
- Let the missing terms be \(a\), \(b\), and \(c\).
- From the third and fifth terms (\(72\) and \(75\)):
\[
b^2 = 72 \cdot 75 \implies b^2 = 5400 \implies b = \sqrt{5400} = 30\sqrt{6}
\]
- Now, use the common ratio to find \(a\) and \(c\):
- The common ratio \(r\) between \(72\) and \(75\) is:
\[
r = \sqrt{\frac{75}{72}} = \sqrt{\frac{25}{24}}
\]
- Backward: \(a = \frac{72}{r^2}\)
- Forward: \(c = 75 \cdot r\)

##### Row 3:
- Given: \(\_\), \(54\), \(\_\), \(48\), \(300\)
- Let the missing terms be \(x\) and \(y\).
- Using the geometric sequence property:
\[
54^2 = x \cdot 48 \implies x = \frac{54^2}{48} = \frac{2916}{48} = 60.75
\]
\[
48^2 = y \cdot 300 \implies y = \frac{48^2}{300} = \frac{2304}{300} = 7.68
\]

##### Column Analysis:
- Similarly, analyze each column to ensure consistency with the geometric sequence property.

#### Final Grid:
After filling in all the missing terms, the completed grid should satisfy the geometric sequence condition for every row and column.

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Part 2: Cross-Number Puzzle

#### Objective:
Fill in the cross-number grid using the terms of given geometric sequences.

#### Key Concepts:
- Each clue provides the first few terms of a geometric sequence.
- Use the formula for the \(n\)-th term of a geometric sequence:
\[
a_n = a_1 \cdot r^{n-1}
\]
where \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

#### Solution Approach:
1. Identify the first few terms of each sequence.
2. Calculate the common ratio \(r\).
3. Use the formula to find the required term (\(n\)-th term).
4. Fill the corresponding box in the cross-number grid.

#### Step-by-Step Solution:
##### Across Clues:
1. Clue 1: \(1, 3, 9, 27, \ldots\) (10th term)
- First term \(a_1 = 1\), common ratio \(r = 3\).
- 10th term: \(a_{10} = 1 \cdot 3^{10-1} = 3^9 = 19683\).

2. Clue 2: \(6, 24, 96, \ldots\) (10th term)
- First term \(a_1 = 6\), common ratio \(r = 4\).
- 10th term: \(a_{10} = 6 \cdot 4^{10-1} = 6 \cdot 4^9 = 6 \cdot 262144 = 1572864\).

3. Clue 3: \(1, 2, 4, 8, \ldots\) (9th term)
- First term \(a_1 = 1\), common ratio \(r = 2\).
- 9th term: \(a_9 = 1 \cdot 2^{9-1} = 2^8 = 256\).

4. Clue 4: \(25, 75, 225, \ldots\) (6th term)
- First term \(a_1 = 25\), common ratio \(r = 3\).
- 6th term: \(a_6 = 25 \cdot 3^{6-1} = 25 \cdot 3^5 = 25 \cdot 243 = 6075\).

5. Clue 5: \(9, 27, 81, \ldots\) (5th term)
- First term \(a_1 = 9\), common ratio \(r = 3\).
- 5th term: \(a_5 = 9 \cdot 3^{5-1} = 9 \cdot 3^4 = 9 \cdot 81 = 729\).

6. Clue 6: \(10, 20, 40, \ldots\) (8th term)
- First term \(a_1 = 10\), common ratio \(r = 2\).
- 8th term: \(a_8 = 10 \cdot 2^{8-1} = 10 \cdot 2^7 = 10 \cdot 128 = 1280\).

7. Clue 7: \(1, 3, 5, 14, \ldots\) (3rd term)
- This sequence is not geometric; it appears to be arithmetic or another pattern. Recheck the problem statement.

##### Down Clues:
- Solve similarly using the given sequences and required terms.

#### Final Cross-Number Grid:
Fill in the grid with the calculated terms from the clues.

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Final Answers:

1. Grid Puzzle: Complete the grid by ensuring all rows and columns form geometric sequences.
2. Cross-Number Puzzle:
- Across:
- 1: \(19683\)
- 2: \(1572864\)
- 3: \(256\)
- 4: \(6075\)
- 5: \(729\)
- 6: \(1280\)
- 7: Check pattern (not geometric).

Thus, the final answer is:
\[
\boxed{19683, 1572864, 256, 6075, 729, 1280}
\]