In a set of four number, the first three are in GP & the last three are in A.P. with common difference 6. If the first number is the same as the fourth, find the four numbers.

Answer

To solve the problem, we need to find four numbers such that the first three numbers are in a geometric progression (GP) and the last three numbers are in an arithmetic progression (AP) with a common difference of 6. Additionally, the first number is equal to the fourth number.

Let's denote the four numbers as follows:
- First number: \( A \)
- Second number: \( AR \) (where \( R \) is the common ratio of the GP)
- Third number: \( AR^2 \)
- Fourth number: \( B \)

Since the last three numbers are in AP with a common difference of 6, we can express them as:
- Third number: \( B - 6 \)
- Fourth number: \( B \)
- Fifth number: \( B + 6 \)

According to the problem, we have:
1. \( A = B + 6 \) (since the first number is equal to the fourth number)
2. \( AR = B - 6 \) (the second number in GP is equal to the third number in AP)
3. \( AR^2 = B \) (the third number in GP is equal to the fourth number in AP)

Now, we can solve these equations step by step.

### Step 1: Set up the equations
From the information given, we have:
1. \( A = B + 6 \) (Equation 1)
2. \( AR = B - 6 \) (Equation 2)
3. \( AR^2 = B \) (Equation 3)

### Step 2: Substitute Equation 1 into Equations 2 and 3
Substituting \( A \) from Equation 1 into Equations 2 and 3:
- From Equation 2:
\[ (B + 6)R = B - 6 \]
- From Equation 3:
\[ (B + 6)R^2 = B \]

### Step 3: Solve for \( R \) from Equation 2
Rearranging Equation 2:
\[ BR + 6R = B - 6 \]
\[ BR - B = -6 - 6R \]
\[ B(R - 1) = -6R - 6 \]
\[ B = \frac{-6(R + 1)}{R - 1} \quad \text{(Equation 4)} \]

### Step 4: Substitute \( B \) into Equation 3
Substituting Equation 4 into Equation 3:
\[ (B + 6)R^2 = B \]
Substituting for \( B \):
\[ \left(\frac{-6(R + 1)}{R - 1} + 6\right)R^2 = \frac{-6(R + 1)}{R - 1} \]

### Step 5: Simplify and solve for \( R \)
This equation can be simplified and solved for \( R \). After simplification, we find:
\[ R = -\frac{1}{2} \quad \text{(after solving the quadratic equation)} \]

### Step 6: Find \( B \) using \( R \)
Now substituting \( R \) back into Equation 4 to find \( B \):
\[ B = \frac{-6\left(-\frac{1}{2} + 1\right)}{-\frac{1}{2} - 1} = \frac{-6\left(\frac{1}{2}\right)}{-\frac{3}{2}} = 4 \]

### Step 7: Find \( A \)
Using \( B = 4 \) in Equation 1:
\[ A = B + 6 = 4 + 6 = 10 \]

### Step 8: Find the four numbers
Now we can find the four numbers:
- First number \( A = 10 \)
- Second number \( AR = 10 \times -\frac{1}{2} = -5 \)
- Third number \( AR^2 = 10 \times \left(-\frac{1}{2}\right)^2 = 10 \times \frac{1}{4} = 2.5 \)
- Fourth number \( B = 4 \)

Thus, the four numbers are:
1. \( 10 \)
2. \( -5 \)
3. \( 2.5 \)
4. \( 4 \)

### Final Answer
The four numbers are \( 10, -5, 2.5, 4 \).