Insert 3 geometric means between 4 and `(1)/(4).`

Explanation

To insert 3 geometric means between 4 and \( \frac{1}{4} \), we can follow these steps:

### Step 1: Identify the terms in the geometric progression (GP)
Let the first term \( a = 4 \) and the fifth term \( a_5 = \frac{1}{4} \). We need to find the three geometric means, which we will denote as \( a_2, a_3, \) and \( a_4 \).

### Step 2: Use the formula for the nth term of a GP
The formula for the nth term of a geometric progression is given by:
\[ a_n = a \cdot r^{n-1} \]
where \( r \) is the common ratio.

### Step 3: Set up the equation for the fifth term
Since \( a_5 = \frac{1}{4} \), we can write:
\[ a_5 = a \cdot r^{5-1} = 4 \cdot r^4 \]
So, we have:
\[ 4 \cdot r^4 = \frac{1}{4} \]

### Step 4: Solve for \( r^4 \)
To isolate \( r^4 \), divide both sides by 4:
\[ r^4 = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

### Step 5: Find \( r \)
Now, take the fourth root of both sides:
\[ r = \pm \sqrt[4]{\frac{1}{16}} = \pm \frac{1}{2} \]

### Step 6: Calculate the geometric means for both cases of \( r \)

#### Case 1: \( r = \frac{1}{2} \)
1. \( a_2 = a \cdot r = 4 \cdot \frac{1}{2} = 2 \)
2. \( a_3 = a \cdot r^2 = 4 \cdot \left(\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \)
3. \( a_4 = a \cdot r^3 = 4 \cdot \left(\frac{1}{2}\right)^3 = 4 \cdot \frac{1}{8} = \frac{1}{2} \)

#### Case 2: \( r = -\frac{1}{2} \)
1. \( a_2 = a \cdot r = 4 \cdot \left(-\frac{1}{2}\right) = -2 \)
2. \( a_3 = a \cdot r^2 = 4 \cdot \left(-\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \)
3. \( a_4 = a \cdot r^3 = 4 \cdot \left(-\frac{1}{2}\right)^3 = 4 \cdot -\frac{1}{8} = -\frac{1}{2} \)

### Final Results
Thus, the three geometric means between 4 and \( \frac{1}{4} \) are:
1. For \( r = \frac{1}{2} \): \( 2, 1, \frac{1}{2} \)
2. For \( r = -\frac{1}{2} \): \( -2, 1, -\frac{1}{2} \)