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.