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 geometric progression (G.P.) given two conditions:

1. The product of the three numbers is 27.
2. The sum of the products of the numbers taken in pairs is 39.

Let's denote the three consecutive numbers in G.P. as:
- \( \frac{a}{r} \) (first term),
- \( a \) (second term),
- \( ar \) (third term).

### Step 1: Set up the equation for the product
According to the first condition, we have:
\[ \frac{a}{r} \cdot a \cdot ar = 27 \]
This simplifies to:
\[ \frac{a^3}{r} = 27 \]
Multiplying both sides by \( r \):
\[ a^3 = 27r \]

### Step 2: Set up the equation for the sum of products taken in pairs
According to the second condition, we have:
\[ \frac{a}{r} \cdot a + a \cdot ar + \frac{a}{r} \cdot ar = 39 \]
This simplifies to:
\[ \frac{a^2}{r} + a^2 + \frac{a^2 r}{r} = 39 \]
This further simplifies to:
\[ \frac{a^2}{r} + a^2 + a^2 = 39 \]
Combining the terms gives:
\[ \frac{a^2}{r} + 2a^2 = 39 \]
Factoring out \( a^2 \):
\[ a^2 \left( \frac{1}{r} + 2 \right) = 39 \]

### Step 3: Substitute \( a^3 \) into the second equation
From \( a^3 = 27r \), we can express \( a^2 \) as:
\[ a^2 = \frac{(27r)^{2/3}}{r^{2/3}} = 27^{2/3} r^{1/3} = 9r^{1/3} \]
Substituting \( a^2 \) into the equation:
\[ 9r^{1/3} \left( \frac{1}{r} + 2 \right) = 39 \]
Distributing gives:
\[ 9 \left( r^{-2/3} + 2r^{1/3} \right) = 39 \]
Dividing by 9:
\[ r^{-2/3} + 2r^{1/3} = \frac{39}{9} = \frac{13}{3} \]

### Step 4: Multiply through by \( r^{2/3} \)
Let \( x = r^{1/3} \), then \( r^{2/3} = x^2 \):
\[ 1 + 2x^3 = \frac{13}{3} x^2 \]
Rearranging gives:
\[ 2x^3 - \frac{13}{3} x^2 + 1 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 6x^3 - 13x^2 + 3 = 0 \]

### Step 5: Solve the cubic equation
Using the Rational Root Theorem or synthetic division, we can find that \( x = 3 \) is a root. Thus, we can factor:
\[ 6x^3 - 13x^2 + 3 = (x - 3)(6x^2 + 5x - 1) \]
Now we can solve for \( x \) using the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} = \frac{-5 \pm \sqrt{25 + 24}}{12} = \frac{-5 \pm 7}{12} \]
This gives us:
\[ x = \frac{2}{12} = \frac{1}{6} \quad \text{or} \quad x = \frac{-12}{12} = -1 \]

### Step 6: Find values of \( r \)
Since \( x = r^{1/3} \):
1. If \( x = 3 \), then \( r = 27 \).
2. If \( x = \frac{1}{6} \), then \( r = \frac{1}{216} \).

### Step 7: Calculate the values of \( a \)
Using \( a^3 = 27r \):
1. For \( r = 27 \):
\[ a^3 = 27 \cdot 27 = 729 \implies a = 9 \]
The numbers are \( \frac{9}{27}, 9, 9 \cdot 27 = 1, 9, 243 \).

2. For \( r = \frac{1}{216} \):
\[ a^3 = 27 \cdot \frac{1}{216} = \frac{27}{216} = \frac{1}{8} \implies a = \frac{1}{2} \]
The numbers are \( \frac{1/2}{1/6}, \frac{1}{2}, \frac{1/2}{1/6} = 3, \frac{1}{2}, \frac{1}{12} \).

### Final Answer
Thus, the two sets of numbers are:
1. \( 1, 3, 9 \)
2. \( 9, 3, 1 \)