Four numbers are in G.P. The sum of first two numbers is 4 and the sum of last two numbers is 36. Find the numbers.

Explanation

To solve the problem, we need to find four numbers in a geometric progression (G.P.) given that the sum of the first two numbers is 4 and the sum of the last two numbers is 36. Let's denote the four numbers as \( a/r^2, a/r, a, ar \), where \( a \) is the first term and \( r \) is the common ratio.

### Step-by-Step Solution:

1. **Define the Numbers**:
Let the four numbers be:
\[ \frac{a}{r^2}, \frac{a}{r}, a, ar \]

2. **Set Up the Equations**:
From the problem, we have the following two equations based on the sums given:
- The sum of the first two numbers:
\[ \frac{a}{r^2} + \frac{a}{r} = 4 \]
- The sum of the last two numbers:
\[ a + ar = 36 \]

3. **Simplify the First Equation**:
Factor out \( \frac{a}{r^2} \) from the first equation:
\[ \frac{a}{r^2} \left( 1 + r \right) = 4 \]
This can be rewritten as:
\[ a \left( \frac{1 + r}{r^2} \right) = 4 \]

4. **Simplify the Second Equation**:
Factor out \( a \) from the second equation:
\[ a(1 + r) = 36 \]
Thus, we can express \( a \) in terms of \( r \):
\[ a = \frac{36}{1 + r} \]

5. **Substitute \( a \) into the First Equation**:
Substitute \( a \) from the second equation into the first equation:
\[ \frac{36}{1 + r} \cdot \frac{1 + r}{r^2} = 4 \]
The \( 1 + r \) cancels out:
\[ \frac{36}{r^2} = 4 \]

6. **Solve for \( r^2 \)**:
Multiply both sides by \( r^2 \):
\[ 36 = 4r^2 \]
Divide both sides by 4:
\[ r^2 = 9 \]
Therefore, \( r = 3 \) or \( r = -3 \).

7. **Find \( a \) for Each Case**:
- **Case 1**: If \( r = 3 \):
\[ a = \frac{36}{1 + 3} = \frac{36}{4} = 9 \]
- **Case 2**: If \( r = -3 \):
\[ a = \frac{36}{1 - 3} = \frac{36}{-2} = -18 \]

8. **Calculate the Numbers**:
- **For \( r = 3 \) and \( a = 9 \)**:
\[ \frac{9}{3^2} = 1, \quad \frac{9}{3} = 3, \quad 9, \quad 9 \cdot 3 = 27 \]
The numbers are \( 1, 3, 9, 27 \).

- **For \( r = -3 \) and \( a = -18 \)**:
\[ \frac{-18}{(-3)^2} = -2, \quad \frac{-18}{-3} = 6, \quad -18, \quad -18 \cdot (-3) = 54 \]
The numbers are \( -2, 6, -18, 54 \).

### Final Answer:
The two sets of numbers are:
1. \( 1, 3, 9, 27 \)
2. \( -2, 6, -18, 54 \)