The sum of four numbers in G.P is 60 and the arithmetic mean of the first and the last numbers is 18 . Find the numbers .

Answer

To solve the problem, we need to find four numbers in Geometric Progression (G.P.) given that their sum is 60 and the arithmetic mean of the first and the last numbers is 18.

### Step-by-Step Solution:

1. **Define the Numbers**:
Let the four numbers in G.P. be \( a, ar, ar^2, ar^3 \).

2. **Set Up the Equations**:
According to the problem, the sum of these numbers is given by:
\[ a + ar + ar^2 + ar^3 = 60 \]
We can factor out \( a \):
\[ a(1 + r + r^2 + r^3) = 60 \]
This can be rewritten as:
\[ a(1 + r)(1 + r^2) = 60 \quad \text{(Equation 1)} \]

3. **Use the Arithmetic Mean Condition**:
The arithmetic mean of the first and last numbers is given by:
\[ \frac{a + ar^3}{2} = 18 \]
Simplifying this gives:
\[ a + ar^3 = 36 \]
Factoring out \( a \):
\[ a(1 + r^3) = 36 \quad \text{(Equation 2)} \]

4. **Divide Equation 2 by Equation 1**:
Dividing Equation 2 by Equation 1 gives:
\[ \frac{a(1 + r^3)}{a(1 + r)(1 + r^2)} = \frac{36}{60} \]
Simplifying this results in:
\[ \frac{1 + r^3}{(1 + r)(1 + r^2)} = \frac{3}{5} \]

5. **Cross Multiply**:
Cross-multiplying gives:
\[ 5(1 + r^3) = 3(1 + r)(1 + r^2) \]
Expanding both sides:
\[ 5 + 5r^3 = 3(1 + r + r^2 + r^3) \]
This simplifies to:
\[ 5 + 5r^3 = 3 + 3r + 3r^2 + 3r^3 \]
Rearranging gives:
\[ 2r^3 - 3r^2 - 3r + 2 = 0 \]

6. **Solve the Cubic Equation**:
We can factor this cubic equation. Let's try \( r = 2 \):
\[ 2(2^3) - 3(2^2) - 3(2) + 2 = 0 \]
This means \( r = 2 \) is a root. We can factor out \( (r - 2) \):
\[ 2r^3 - 3r^2 - 3r + 2 = (r - 2)(2r^2 + r - 1) \]

7. **Find Other Roots**:
Now we need to solve \( 2r^2 + r - 1 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us \( r = \frac{1}{2} \) and \( r = -1 \).

8. **Calculate \( a \) for Each \( r \)**:
- For \( r = 2 \):
\[ a(1 + 8) = 36 \implies 9a = 36 \implies a = 4 \]
The numbers are \( 4, 8, 16, 32 \).

- For \( r = \frac{1}{2} \):
\[ a(1 + \frac{1}{8}) = 36 \implies \frac{9}{8}a = 36 \implies a = 32 \]
The numbers are \( 32, 16, 8, 4 \).

### Conclusion:
The four numbers in G.P. are \( 4, 8, 16, 32 \) or \( 32, 16, 8, 4 \).