The product of three numbers in G.P. is 64 and their sum is 14. Find the numbers.

Explanation

To solve the problem, we need to find three numbers in geometric progression (G.P.) whose product is 64 and whose sum is 14. Let's denote the three numbers as \( a/r \), \( a \), and \( ar \), where \( a \) is the middle term and \( r \) is the common ratio.

### Step 1: Set up the equations
Given:
1. The product of the numbers:
\[ \frac{a}{r} \cdot a \cdot ar = 64 \]
This simplifies to:
\[ a^3 = 64 \]

2. The sum of the numbers:
\[ \frac{a}{r} + a + ar = 14 \]

### Step 2: Solve for \( a \)
From the product equation:
\[ a^3 = 64 \]
Taking the cube root of both sides gives:
\[ a = 4 \]

### Step 3: Substitute \( a \) into the sum equation
Now substitute \( a = 4 \) into the sum equation:
\[ \frac{4}{r} + 4 + 4r = 14 \]
To eliminate the fraction, multiply the entire equation by \( r \):
\[ 4 + 4r + 4r^2 = 14r \]
Rearranging gives:
\[ 4r^2 - 10r + 4 = 0 \]

### Step 4: Solve the quadratic equation
Now we will solve the quadratic equation \( 4r^2 - 10r + 4 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = -10 \), and \( c = 4 \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \]
Calculating the discriminant:
\[ r = \frac{10 \pm \sqrt{100 - 64}}{8} = \frac{10 \pm \sqrt{36}}{8} = \frac{10 \pm 6}{8} \]
This gives us two possible values for \( r \):
1. \( r = \frac{16}{8} = 2 \)
2. \( r = \frac{4}{8} = \frac{1}{2} \)

### Step 5: Find the three numbers
Now we can find the three numbers for each value of \( r \):

**Case 1: \( r = 2 \)**
- First number: \( \frac{4}{2} = 2 \)
- Second number: \( 4 \)
- Third number: \( 4 \cdot 2 = 8 \)

So, the numbers are \( 2, 4, 8 \).

**Case 2: \( r = \frac{1}{2} \)**
- First number: \( \frac{4}{\frac{1}{2}} = 8 \)
- Second number: \( 4 \)
- Third number: \( 4 \cdot \frac{1}{2} = 2 \)

So, the numbers are \( 8, 4, 2 \).

### Final Answer
The three numbers in G.P. are \( 2, 4, 8 \) or \( 8, 4, 2 \).