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.