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

Explanation

To solve the problem of finding three numbers in a Geometric Progression (G.P.) whose product is 64 and sum is 14, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Numbers**:
Let the three numbers in G.P. be \( \frac{A}{R}, A, AR \), where \( A \) is the middle term and \( R \) is the common ratio.

2. **Set Up the Product Equation**:
The product of the three numbers is given as:
\[ \frac{A}{R} \cdot A \cdot AR = 64 \]
Simplifying this, we have:
\[ \frac{A^3}{R} = 64 \]
Therefore, we can express this as:
\[ A^3 = 64R \]

3. **Set Up the Sum Equation**:
The sum of the three numbers is given as:
\[ \frac{A}{R} + A + AR = 14 \]
Multiplying through by \( R \) to eliminate the fraction gives:
\[ A + AR + A R^2 = 14R \]
This can be rearranged to:
\[ A(1 + R + R^2) = 14R \]

4. **Substituting for A**:
From the product equation, we can express \( A \) in terms of \( R \):
\[ A = \sqrt[3]{64R} = 4\sqrt[3]{R} \]
Substitute this value of \( A \) into the sum equation:
\[ 4\sqrt[3]{R}(1 + R + R^2) = 14R \]

5. **Simplifying the Equation**:
Dividing both sides by 2 gives:
\[ 2\sqrt[3]{R}(1 + R + R^2) = 7R \]
Rearranging leads to:
\[ 2\sqrt[3]{R} + 2\sqrt[3]{R}R + 2\sqrt[3]{R}R^2 = 7R \]

6. **Letting \( x = \sqrt[3]{R} \)**:
Substituting \( R = x^3 \) gives:
\[ 2x(1 + x^3 + x^6) = 7x^3 \]
Simplifying this leads to:
\[ 2x + 2x^4 + 2x^7 = 7x^3 \]
Rearranging gives:
\[ 2x^7 + 2x^4 - 7x^3 + 2x = 0 \]

7. **Factoring the Polynomial**:
This can be factored or solved using the Rational Root Theorem or synthetic division. After solving, we find:
\[ R = 2 \quad \text{or} \quad R = \frac{1}{2} \]

8. **Finding A and the Numbers**:
If \( R = 2 \):
\[ A = 4 \]
The numbers are:
\[ \frac{4}{2}, 4, 4 \cdot 2 = 2, 4, 8 \]

If \( R = \frac{1}{2} \):
\[ A = 4 \]
The numbers are:
\[ \frac{4}{\frac{1}{2}}, 4, 4 \cdot \frac{1}{2} = 8, 4, 2 \]

9. **Conclusion**:
The three numbers in G.P. are \( 2, 4, 8 \).