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

Answer

To solve the problem step by step, we will denote the four numbers as \( a, b, c, d \).

### Step 1: Set up the relationships
We know that:
- The first three numbers \( a, b, c \) are in a geometric progression (G.P.).
- The last three numbers \( b, c, d \) are in an arithmetic progression (A.P.).
- The difference in the A.P. is given as 6.
- The first number \( a \) is the same as the fourth number \( d \).

### Step 2: Express the A.P. condition
Since \( b, c, d \) are in A.P., we can express the relationship as:
\[ c - b = d - c = 6 \]
From this, we can derive:
1. \( c = b + 6 \)
2. \( d = c + 6 = (b + 6) + 6 = b + 12 \)

### Step 3: Use the condition that \( a = d \)
Given that \( a = d \), we can substitute \( d \):
\[ a = b + 12 \]

### Step 4: Rewrite the numbers
Now we can express all four numbers in terms of \( b \):
- \( a = b + 12 \)
- \( b = b \)
- \( c = b + 6 \)
- \( d = b + 12 \)

So, the four numbers are:
\[ (b + 12), b, (b + 6), (b + 12) \]

### Step 5: Set up the G.P. condition
For \( a, b, c \) to be in G.P., we need:
\[ \frac{b}{a} = \frac{c}{b} \]
Substituting the values we have:
\[ \frac{b}{b + 12} = \frac{b + 6}{b} \]

### Step 6: Cross-multiply
Cross-multiplying gives:
\[ b^2 = (b + 6)(b + 12) \]

### Step 7: Expand and simplify
Expanding the right side:
\[ b^2 = b^2 + 12b + 6b + 72 \]
This simplifies to:
\[ b^2 = b^2 + 18b + 72 \]
Subtract \( b^2 \) from both sides:
\[ 0 = 18b + 72 \]

### Step 8: Solve for \( b \)
Rearranging gives:
\[ 18b = -72 \implies b = -4 \]

### Step 9: Find the other numbers
Now substituting \( b = -4 \) back into our expressions:
- \( a = b + 12 = -4 + 12 = 8 \)
- \( c = b + 6 = -4 + 6 = 2 \)
- \( d = b + 12 = -4 + 12 = 8 \)

### Final answer
The four numbers are:
\[ 8, -4, 2, 8 \]