1.What is the 5th term of the geometric sequence 3/20,3/2,15,...
2.Find the sixth term of a geometric sequence where the second term is 6 and the common ratio is 2
3.In the geometric sequence 6,12,24,48, which term is 768?
4.The second term of the geometric sequence is 3/4 and the fourth term is 3. What is the first term?
5.Find the two geometric means between -32 and 4.

Let's go through each problem one by one.

### 1. Find the 5th term of the geometric sequence: \( \frac{3}{20}, \frac{3}{2}, 15, \dots \)
We need to find the common ratio \(r\). 

The common ratio is:
\[
r = \frac{\frac{3}{2}}{\frac{3}{20}} = \frac{3}{2} \times \frac{20}{3} = 10
\]

Now, to find the 5th term, we use the formula for the \(n\)-th term of a geometric sequence:
\[
a_n = a_1 \cdot r^{n-1}
\]
Here, \(a_1 = \frac{3}{20}\), \(r = 10\), and \(n = 5\).
\[
a_5 = \frac{3}{20} \cdot 10^{4} = \frac{3}{20} \cdot 10000 = \frac{30000}{20} = 1500
\]
So, the 5th term is **1500**.

### 2. Find the sixth term of a geometric sequence where the second term is 6 and the common ratio is 2.

The formula for the \(n\)-th term of a geometric sequence is:
\[
a_n = a_1 \cdot r^{n-1}
\]
We are given \(r = 2\) and \(a_2 = 6\). To find \(a_1\), use the second term formula:
\[
6 = a_1 \cdot 2^{1} \implies a_1 = 6 / 2 = 3
\]

Now, to find the 6th term:
\[
a_6 = a_1 \cdot 2^{5} = 3 \cdot 32 = 96
\]
So, the 6th term is **96**.

### 3. In the geometric sequence \(6, 12, 24, 48, \dots\), which term is 768?

We know the common ratio \(r = \frac{12}{6} = 2\), and the first term \(a_1 = 6\).

Using the formula for the \(n\)-th term:
\[
a_n = a_1 \cdot r^{n-1}
\]
We set \(a_n = 768\):
\[
768 = 6 \cdot 2^{n-1}
\]
\[
\frac{768}{6} = 2^{n-1} \implies 128 = 2^{n-1}
\]
\[
128 = 2^7 \implies n-1 = 7 \implies n = 8
\]
So, **768** is the **8th term**.

### 4. The second term of the geometric sequence is \( \frac{3}{4} \) and the fourth term is 3. What is the first term?

Let the first term be \(a_1\) and the common ratio be \(r\). We know:
\[
a_2 = a_1 \cdot r = \frac{3}{4}
\]
\[
a_4 = a_1 \cdot r^3 = 3
\]

From \(a_2 = a_1 \cdot r = \frac{3}{4}\), we get:
\[
r = \frac{3}{4a_1}
\]

Substitute this into \(a_4 = a_1 \cdot r^3 = 3\):
\[
a_1 \cdot \left(\frac{3}{4a_1}\right)^3 = 3
\]
\[
a_1 \cdot \frac{27}{64a_1^3} = 3
\]
\[
\frac{27}{64a_1^2} = 3 \implies 27 = 192a_1^2 \implies a_1^2 = \frac{27}{192} = \frac{9}{64} \implies a_1 = \frac{3}{8}
\]
So, the first term is **\( \frac{3}{8} \)**.

### 5. Find the two geometric means between \(-32\) and \(4\).

Let the sequence be \( -32, x_1, x_2, 4 \). Since it's a geometric sequence, the common ratio is constant:
\[
r = \frac{x_1}{-32} = \frac{x_2}{x_1} = \frac{4}{x_2}
\]

From \(r = \frac{x_1}{-32}\) and \(r = \frac{4}{x_2}\), we get:
\[
x_2 = r \cdot x_1 \quad \text{and} \quad r^3 = \frac{4}{-32} = -\frac{1}{8}
\]

Thus, \(r = -\frac{1}{2}\). Now, calculate the geometric means:
\[
x_1 = -32 \cdot \left(-\frac{1}{2}\right) = 16
\]
\[
x_2 = 16 \cdot \left(-\frac{1}{2}\right) = -8
\]

So, the two geometric means are **16** and **-8**.

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Would you like further details on any of these steps?

Here are five related questions to consider:
1. How do you find the sum of a finite geometric series?
2. What happens if the common ratio in a geometric sequence is negative?
3. How do you derive the formula for the \(n\)-th term in a geometric sequence?
4. How would you calculate the sum of an infinite geometric series?
5. Can a geometric sequence have a fractional common ratio?

**Tip**: To verify the correctness of terms in a geometric sequence, always check if the ratio between consecutive terms is constant.

Explore solutions to various geometric sequence problems, including finding specific terms and common ratios. Learn the formula for the nth term of a geometric sequence and practical applications in math. Suitable for high school students.

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