The sum of three numbers in GP. Is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the numbers in GP
Let the three numbers in geometric progression (GP) be:
- \( a \) (the first term)
- \( ar \) (the second term)
- \( ar^2 \) (the third term)

### Step 2: Set up the equation for the sum
According to the problem, the sum of these three numbers is given as:
\[ a + ar + ar^2 = 56 \]
Factoring out \( a \), we have:
\[ a(1 + r + r^2) = 56 \tag{1} \]

### Step 3: Set up the equation for the AP condition
When we subtract 1, 7, and 21 from the three numbers respectively, we get:
- \( a - 1 \)
- \( ar - 7 \)
- \( ar^2 - 21 \)

These numbers must be in arithmetic progression (AP). For three numbers \( x, y, z \) to be in AP, the condition is:
\[ 2y = x + z \]
Applying this to our numbers:
\[ 2(ar - 7) = (a - 1) + (ar^2 - 21) \]
Expanding this gives:
\[ 2ar - 14 = a - 1 + ar^2 - 21 \]
Simplifying, we get:
\[ 2ar - 14 = a + ar^2 - 22 \]
Rearranging this leads to:
\[ ar^2 - 2ar + a - 8 = 0 \tag{2} \]

### Step 4: Substitute \( a \) from equation (1) into equation (2)
From equation (1), we can express \( a \) as:
\[ a = \frac{56}{1 + r + r^2} \]
Substituting this into equation (2):
\[ \frac{56}{1 + r + r^2} r^2 - 2\frac{56}{1 + r + r^2}r + \frac{56}{1 + r + r^2} - 8 = 0 \]
Multiplying through by \( 1 + r + r^2 \) to eliminate the denominator:
\[ 56r^2 - 112r + 56 - 8(1 + r + r^2) = 0 \]
Expanding gives:
\[ 56r^2 - 112r + 56 - 8 - 8r - 8r^2 = 0 \]
Combining like terms results in:
\[ 48r^2 - 120r + 48 = 0 \]
Dividing the entire equation by 12 simplifies it to:
\[ 4r^2 - 10r + 4 = 0 \tag{3} \]

### Step 5: Solve the quadratic equation (3)
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -10, c = 4 \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \]
Calculating the discriminant:
\[ r = \frac{10 \pm \sqrt{100 - 64}}{8} = \frac{10 \pm \sqrt{36}}{8} = \frac{10 \pm 6}{8} \]
This gives us:
\[ r = \frac{16}{8} = 2 \quad \text{or} \quad r = \frac{4}{8} = \frac{1}{2} \]

### Step 6: Find \( a \) for both values of \( r \)
1. **When \( r = 2 \)**:
\[ a = \frac{56}{1 + 2 + 4} = \frac{56}{7} = 8 \]
The numbers are:
\[ 8, 16, 32 \]

2. **When \( r = \frac{1}{2} \)**:
\[ a = \frac{56}{1 + \frac{1}{2} + \frac{1}{4}} = \frac{56}{\frac{7}{4}} = 32 \]
The numbers are:
\[ 32, 16, 8 \]

### Conclusion
The three numbers in GP are \( 8, 16, 32 \) or \( 32, 16, 8 \).

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