The sum of three consecutive numbers in a G.P. is 56. If we subtract 1, 7, 21 respectively from these numbers, the new numbers form an A.P. Find the numbers.

Explanation

To solve the problem step by step, we will follow the same reasoning as outlined in the video transcript.

### Step 1: Define the numbers in the G.P.
Let the three consecutive numbers in a geometric progression (G.P.) be:
- First number: \( a \)
- Second number: \( ar \)
- Third number: \( ar^2 \)

### Step 2: Set up the equation for the sum
According to the problem, the sum of these three numbers is 56. Therefore, we can write the equation:
\[ a + ar + ar^2 = 56 \]
Factoring out \( a \), we get:
\[ a(1 + r + r^2) = 56 \]
Let’s call this **Equation 1**.

### Step 3: Set up the condition for A.P.
We are told that if we subtract 1, 7, and 21 from the first, second, and third numbers respectively, the resulting numbers form an arithmetic progression (A.P.). The new numbers will be:
- First number: \( a - 1 \)
- Second number: \( ar - 7 \)
- Third number: \( ar^2 - 21 \)

For these numbers to be in A.P., the condition is:
\[ 2(ar - 7) = (a - 1) + (ar^2 - 21) \]
Expanding this gives:
\[ 2ar - 14 = a + ar^2 - 22 \]
Rearranging this, we have:
\[ 2ar - ar^2 - a + 8 = 0 \]
Let’s call this **Equation 2**.

### Step 4: Rearranging Equation 2
Rearranging Equation 2 gives:
\[ -ar^2 + 2ar - a + 8 = 0 \]
This can be rewritten as:
\[ a - 2ar + ar^2 = 8 \]

### Step 5: Substitute \( a \) from Equation 1 into Equation 2
From Equation 1, we can express \( a \):
\[ a = \frac{56}{1 + r + r^2} \]
Substituting this into the rearranged Equation 2:
\[ \frac{56}{1 + r + r^2} - 2r\left(\frac{56}{1 + r + r^2}\right) + r^2\left(\frac{56}{1 + r + r^2}\right) = 8 \]
Multiplying through by \( 1 + r + r^2 \) to eliminate the fraction:
\[ 56 - 112r + 56r^2 = 8(1 + r + r^2) \]
Expanding the right side:
\[ 56 - 112r + 56r^2 = 8 + 8r + 8r^2 \]
Rearranging gives:
\[ 48r^2 - 120r + 48 = 0 \]
Dividing through by 24 simplifies this to:
\[ 2r^2 - 5r + 2 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ r = \frac{5 \pm \sqrt{25 - 16}}{4} \]
\[ r = \frac{5 \pm 3}{4} \]
This gives us two possible values for \( r \):
\[ r = 2 \quad \text{or} \quad r = \frac{1}{2} \]

### Step 7: Find the values of \( a \)
1. **For \( r = 2 \)**:
\[ a(1 + 2 + 4) = 56 \implies 7a = 56 \implies a = 8 \]
The numbers are \( 8, 16, 32 \).

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

### Final Answer
The three numbers in G.P. are \( 8, 16, 32 \) or \( 32, 16, 8 \).