To solve the problem, we need to find three numbers in geometric progression (G.P.) whose sum is 56, and when we subtract 1, 7, and 21 from these numbers respectively, the results form an arithmetic progression (A.P.).
### Step-by-Step Solution:
1. **Define the Numbers in G.P.**:
Let the three numbers in G.P. be \( a \), \( ar \), and \( ar^2 \).
2. **Set Up the Equation for the Sum**:
According to the problem, the sum of these numbers is given by:
\[
a + ar + ar^2 = 56
\]
We can factor this equation:
\[
a(1 + r + r^2) = 56 \quad \text{(Equation 1)}
\]
3. **Set Up the Condition for A.P.**:
The problem states that if we subtract 1, 7, and 21 from the numbers respectively, we obtain an A.P. This gives us:
- First term: \( a - 1 \)
- Second term: \( ar - 7 \)
- Third term: \( ar^2 - 21 \)
For these to be in A.P., the condition is:
\[
2(ar - 7) = (a - 1) + (ar^2 - 21)
\]
Simplifying this:
\[
2ar - 14 = a - 1 + ar^2 - 21
\]
Rearranging gives:
\[
2ar - a + 7 = ar^2 \quad \text{(Equation 2)}
\]
4. **Rearranging Equation 2**:
We can rearrange Equation 2 to:
\[
ar^2 - 2ar + a - 7 = 0
\]
5. **Substituting \( a \) from Equation 1**:
From Equation 1, we can express \( a \) as:
\[
a = \frac{56}{1 + r + r^2}
\]
Substitute \( a \) into Equation 2:
\[
ar^2 - 2ar + \frac{56}{1 + r + r^2} - 7 = 0
\]
6. **Multiply through by \( 1 + r + r^2 \)** to eliminate the fraction:
\[
ar^2(1 + r + r^2) - 2ar(1 + r + r^2) + 56 - 7(1 + r + r^2) = 0
\]
7. **Expanding and simplifying**:
This will yield a quadratic equation in terms of \( r \). After simplification, we will arrive at:
\[
6r^2 - 15r + 6 = 0
\]
8. **Solving the Quadratic Equation**:
We can use the quadratic formula:
\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 6 \cdot 6}}{2 \cdot 6}
\]
This simplifies to:
\[
r = \frac{15 \pm 3}{12}
\]
Thus, we find:
\[
r = 2 \quad \text{or} \quad r = \frac{1}{2}
\]
9. **Finding \( a \) for each \( r \)**:
- If \( r = 2 \):
\[
a(1 + 2 + 4) = 56 \implies 7a = 56 \implies a = 8
\]
The numbers are \( 8, 16, 32 \).
- If \( r = \frac{1}{2} \):
\[
a(1 + \frac{1}{2} + \frac{1}{4}) = 56 \implies a \cdot \frac{7}{4} = 56 \implies a = 32
\]
The numbers are \( 32, 16, 8 \).
10. **Conclusion**:
In both cases, the numbers in G.P. are \( 8, 16, 32 \).
### Final Answer:
The three numbers are \( 8, 16, 32 \).
