III. Direction: Insert geometric means in each geometric sequence.
1. 3, _, 8
2. 32, _,38
3. 7, _, _, 56
4. 36, _, _, _, 964
5.-1, _, -9, _, -81

To solve the problems of inserting geometric means into each geometric sequence, we need to understand the concept of geometric sequences. In a geometric sequence, each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

### Problem 1: \(3, \_, 8\)
Let's find the missing term \(x\).
- Here, the sequence has two terms and one missing term. The first term is 3, the last term is 8, and \(x\) is the geometric mean.
- In a geometric sequence, the square of the geometric mean is equal to the product of the first and third terms: 
  \[
  x^2 = 3 \times 8
  \]
  \[
  x^2 = 24 \quad \Rightarrow \quad x = \sqrt{24} = 2\sqrt{6} \approx 4.9
  \]
  Therefore, the sequence is approximately \(3, 4.9, 8\).

### Problem 2: \(32, \_, 38\)
Let's find the missing term \(x\).
- Here, the sequence has two terms and one missing term. The first term is 32, the last term is 38, and \(x\) is the geometric mean.
- Using the same principle:
  \[
  x^2 = 32 \times 38 = 1216
  \]
  \[
  x = \sqrt{1216} \approx 34.86
  \]
  Therefore, the sequence is approximately \(32, 34.86, 38\).

### Problem 3: \(7, \_, \_, 56\)
Let's find the two missing terms \(x_1\) and \(x_2\).
- Here, there are two missing terms, so we have:
  \[
  7, x_1, x_2, 56
  \]
- The common ratio \(r\) can be found by solving:
  \[
  56 = 7r^3 \quad \Rightarrow \quad r^3 = \frac{56}{7} = 8 \quad \Rightarrow \quad r = \sqrt[3]{8} = 2
  \]
- Now, \(x_1 = 7 \times 2 = 14\) and \(x_2 = 14 \times 2 = 28\).
  Therefore, the sequence is \(7, 14, 28, 56\).

### Problem 4: \(36, \_, \_, \_, 964\)
Let's find the three missing terms \(x_1\), \(x_2\), and \(x_3\).
- Here, there are three missing terms, so we have:
  \[
  36, x_1, x_2, x_3, 964
  \]
- The common ratio \(r\) can be found by solving:
  \[
  964 = 36r^4 \quad \Rightarrow \quad r^4 = \frac{964}{36} = 26.78 \quad \Rightarrow \quad r \approx \sqrt[4]{26.78} \approx 2.43
  \]
- Now, \(x_1 = 36 \times 2.43 \approx 87.48\), \(x_2 = 87.48 \times 2.43 \approx 212.56\), and \(x_3 = 212.56 \times 2.43 \approx 516.14\).
  Therefore, the sequence is approximately \(36, 87.48, 212.56, 516.14, 964\).

### Problem 5: \(-1, \_, -9, \_, -81\)
Let's find the two missing terms \(x_1\) and \(x_2\).
- Here, there are two missing terms, so we have:
  \[
  -1, x_1, -9, x_2, -81
  \]
- The common ratio \(r\) can be found by solving:
  \[
  -9 = -1 \times r^2 \quad \Rightarrow \quad r^2 = 9 \quad \Rightarrow \quad r = 3 \text{ or } r = -3
  \]
- Case 1: \(r = 3\):
  \[
  x_1 = -1 \times 3 = -3, \quad x_2 = -9 \times 3 = -27
  \]
  Sequence: \(-1, -3, -9, -27, -81\)

- Case 2: \(r = -3\):
  \[
  x_1 = -1 \times (-3) = 3, \quad x_2 = -9 \times (-3) = 27
  \]
  Sequence: \(-1, 3, -9, 27, -81\)
  
Since we're following the sign pattern, the appropriate solution is \(r = -3\), so the sequence is \(-1, 3, -9, 27, -81\).

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Would you like further details on any of these steps or have any questions? Here are five related questions you might find interesting:

1. How do you find the common ratio in a geometric sequence with more than two missing terms?
2. What is the difference between arithmetic and geometric sequences?
3. Can a geometric sequence have a common ratio of 1 or -1?
4. How do you handle sequences with negative terms when finding geometric means?
5. What happens if a sequence has a non-integer common ratio?

**Tip:** When working with geometric sequences, always ensure that the common ratio remains consistent throughout the sequence to accurately find missing terms.

Explore how to insert geometric means into geometric sequences with detailed solutions for each problem. Learn the formula for finding geometric means in sequences. Suitable for high school students studying geometric sequences and series.

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