To find three geometric means between 16 and 81, we'll use the geometric sequence formula.

The formula for the $n$-th term of a geometric sequence is:

$a_n = a_1 \cdot r^{(n-1)}$

Where:

• $a_1 = 16$ (the first term)
• $a_5 = 81$ (the fifth term, since there are 3 geometric means between them, making the total sequence length 5)
• $r$ is the common ratio, which we need to find.

We know:

$a_5 = a_1 \cdot r^4$

Substitute the known values:

$81 = 16 \cdot r^4$

Now, solve for $r$:

$r^4 = \frac{81}{16} = 5.0625$

Take the fourth root of both sides:

$r = \sqrt[4]{5.0625} \approx 1.5$

Now that we have the common ratio $r = 1.5$, we can find the three geometric means:

1. The second term $a_2 = a_1 \cdot r = 16 \cdot 1.5 = 24$
2. The third term $a_3 = a_2 \cdot r = 24 \cdot 1.5 = 36$
3. The fourth term $a_4 = a_3 \cdot r = 36 \cdot 1.5 = 54$

Thus, the three geometric means between 16 and 81 are:

$\boxed{24}, \boxed{36}, \boxed{54}$

Would you like more details or have any questions?

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Here are five related questions to expand on this topic:

1. How do you find the geometric mean of two numbers?
2. What is the difference between arithmetic and geometric sequences?
3. Can you derive the formula for the $n$-th term of a geometric sequence?
4. How do you solve a geometric sequence problem with more than three means?
5. What happens if one or both of the terms are negative?

Tip: In a geometric sequence, each term is a constant multiple (ratio) of the previous term!