Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Explanation

To find four numbers forming a geometric progression (GP) under the given conditions, let's denote the four terms of the GP as:

1st term = \( a \)
2nd term = \( ar \)
3rd term = \( ar^2 \)
4th term = \( ar^3 \)

Given conditions:
1. The third term is greater than the first term by 9:
\[ ar^2 = a + 9 \quad (1) \]

2. The second term is greater than the fourth term by 18:
\[ ar = ar^3 + 18 \quad (2) \]

### Step 1: Solve equation (1)
From equation (1):
\[ ar^2 - a = 9 \]
Factoring out \( a \):
\[ a(r^2 - 1) = 9 \quad (3) \]

### Step 2: Solve equation (2)
From equation (2):
\[ ar - ar^3 = 18 \]
Factoring out \( ar \):
\[ ar(1 - r^2) = 18 \quad (4) \]

### Step 3: Express \( a \) in terms of \( r \)
From equation (3):
\[ a = \frac{9}{r^2 - 1} \quad (5) \]

### Step 4: Substitute \( a \) in equation (4)
Substituting equation (5) into equation (4):
\[ \frac{9}{r^2 - 1} \cdot r(1 - r^2) = 18 \]
Simplifying:
\[ \frac{9r(1 - r^2)}{r^2 - 1} = 18 \]
This simplifies to:
\[ 9r(1 - r^2) = 18(r^2 - 1) \]
Expanding both sides:
\[ 9r - 9r^3 = 18r^2 - 18 \]
Rearranging gives:
\[ 9r^3 + 18r^2 - 9r - 18 = 0 \]
Dividing the entire equation by 9:
\[ r^3 + 2r^2 - r - 2 = 0 \quad (6) \]

### Step 5: Solve the cubic equation (6)
To solve \( r^3 + 2r^2 - r - 2 = 0 \), we can use the Rational Root Theorem to test possible rational roots. Testing \( r = 1 \):
\[ 1^3 + 2(1^2) - 1 - 2 = 1 + 2 - 1 - 2 = 0 \]
Thus, \( r = 1 \) is a root. We can factor the polynomial:
\[ (r - 1)(r^2 + 3r + 2) = 0 \]
Factoring further:
\[ (r - 1)(r + 1)(r + 2) = 0 \]
Thus, the roots are:
\[ r = 1, r = -1, r = -2 \]

### Step 6: Find corresponding \( a \) values
1. **For \( r = 1 \)**:
- From equation (5):
\[ a = \frac{9}{1^2 - 1} \quad \text{(undefined)} \]

2. **For \( r = -1 \)**:
- From equation (5):
\[ a = \frac{9}{(-1)^2 - 1} \quad \text{(undefined)} \]

3. **For \( r = -2 \)**:
- From equation (5):
\[ a = \frac{9}{(-2)^2 - 1} = \frac{9}{4 - 1} = \frac{9}{3} = 3 \]
- Thus, the terms are:
\[ a = 3, \quad ar = 3(-2) = -6, \quad ar^2 = 3(-2)^2 = 12, \quad ar^3 = 3(-2)^3 = -24 \]

### Final Answer
The four numbers forming the geometric progression are:
\[ 3, -6, 12, -24 \]