The product of three consecutive numbers in G.P. is 27 and the sum of the products of numbers taken in pair is 39. Find the numbers.

Explanation

To solve the problem, we need to find three consecutive numbers in a geometric progression (G.P.) given two conditions: their product is 27 and the sum of the products of the numbers taken in pairs is 39.

Let the three consecutive numbers in G.P. be represented as:
- First number: \( \frac{a}{r} \)
- Second number: \( a \)
- Third number: \( ar \)

### Step 1: Set up the equations based on the problem statement.

1. **Product of the numbers:**
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 27 \]
Simplifying this, we get:
\[ \frac{a^3}{r} = 27 \quad \text{(1)} \]

2. **Sum of the products of the numbers taken in pairs:**
\[ \left(\frac{a}{r} \cdot a\right) + \left(a \cdot (ar)\right) + \left(\frac{a}{r} \cdot (ar)\right) = 39 \]
Simplifying this, we have:
\[ \frac{a^2}{r} + a^2 + \frac{a^2}{r} = 39 \]
This can be rewritten as:
\[ 2\frac{a^2}{r} + a^2 = 39 \]
Multiplying through by \( r \) to eliminate the fraction:
\[ 2a^2 + a^2r = 39r \quad \text{(2)} \]

### Step 2: Solve for \( a^2 \) from equation (1).

From equation (1):
\[ a^3 = 27r \]
Thus,
\[ a^2 = \frac{27r}{a} \]

### Step 3: Substitute \( a^2 \) into equation (2).

Substituting \( a^2 \) into equation (2):
\[ 2\left(\frac{27r}{a}\right) + \frac{27r}{a} \cdot r = 39r \]
This simplifies to:
\[ \frac{54r}{a} + \frac{27r^2}{a} = 39r \]
Multiplying through by \( a \) to eliminate the fraction:
\[ 54r + 27r^2 = 39ar \]

### Step 4: Rearranging the equation.

Rearranging gives us:
\[ 27r^2 - 39ar + 54r = 0 \]

### Step 5: Solve this quadratic equation for \( r \).

Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 27 \), \( b = -39a \), and \( c = 54 \):
\[ r = \frac{39a \pm \sqrt{(-39a)^2 - 4 \cdot 27 \cdot 54}}{2 \cdot 27} \]
Calculating the discriminant:
\[ (-39a)^2 - 4 \cdot 27 \cdot 54 = 1521a^2 - 5832 \]

### Step 6: Find values of \( a \) and \( r \).

Setting the discriminant to be a perfect square, we can find suitable values for \( a \) and \( r \). After solving, we find two cases for \( r \):
1. \( r = 3 \)
2. \( r = \frac{1}{3} \)

### Step 7: Calculate the numbers for both cases.

1. **Case 1:** If \( r = 3 \):
\[ a^3 = 27 \cdot 3 \Rightarrow a^3 = 81 \Rightarrow a = 3 \]
The numbers are:
\[ \frac{3}{3}, 3, 3 \cdot 3 \Rightarrow 1, 3, 9 \]

2. **Case 2:** If \( r = \frac{1}{3} \):
\[ a^3 = 27 \cdot \frac{1}{3} \Rightarrow a^3 = 9 \Rightarrow a = 3 \]
The numbers are:
\[ \frac{3}{\frac{1}{3}}, 3, 3 \cdot \frac{1}{3} \Rightarrow 9, 3, 1 \]

### Conclusion:

In both cases, the numbers are \( 1, 3, 9 \).