To insert 4 geometric means between two numbers, $a_1 = 72$ and $a_6 = \frac{9}{4}$, we first need to understand that these numbers will form a geometric sequence. The general form of a geometric sequence is:

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

Where:

• $a_n$ is the $n$-th term of the sequence.
• $r$ is the common ratio.
• $n$ is the position of the term in the sequence.

Given $a_1 = 72$ and $a_6 = \frac{9}{4}$, we have:

$a_6 = a_1 \cdot r^{(6-1)} = 72 \cdot r^5 = \frac{9}{4}$

We need to solve for $r$:

$r^5 = \frac{\frac{9}{4}}{72} = \frac{9}{4} \times \frac{1}{72} = \frac{9}{288} = \frac{1}{32}$

Taking the 5th root of both sides:

$r = \left(\frac{1}{32}\right)^{\frac{1}{5}}$

Since $32 = 2^5$, we have:

$r = \frac{1}{2}$

Now that we have the common ratio $r = \frac{1}{2}$, we can find the four geometric means between 72 and $\frac{9}{4}$:

1. $a_2 = a_1 \cdot r = 72 \times \frac{1}{2} = 36$
2. $a_3 = a_2 \cdot r = 36 \times \frac{1}{2} = 18$
3. $a_4 = a_3 \cdot r = 18 \times \frac{1}{2} = 9$
4. $a_5 = a_4 \cdot r = 9 \times \frac{1}{2} = 4.5$

Thus, the geometric sequence is $72, 36, 18, 9, 4.5, \frac{9}{4}$.

Summary

The 4 geometric means between 72 and $\frac{9}{4}$ are: $36, 18, 9,$ and $4.5$.

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Related Questions

1. How do you calculate the geometric mean of two numbers?
2. What is the difference between arithmetic and geometric sequences?
3. Can a geometric sequence contain negative numbers?
4. How does the geometric mean relate to exponential growth?
5. How would you insert geometric means between negative numbers?

Tip

When solving for a common ratio in a geometric sequence, always check if the result is a rational or an irrational number, as this impacts the type of sequence formed.