Three number are in A.P. such that their sum is
24 and sum of their squares is 224 . The number are

Options

A. 2, 8, 14 `
B. `4, 8, 12 `
C. 6,8,12
D. 5, 8, 11

Answer

To solve the problem, we need to find three numbers in Arithmetic Progression (A.P.) such that their sum is 24 and the sum of their squares is 224.

### Step-by-Step Solution:

1. **Assume the numbers in A.P.**:
Let the three numbers be \( a - d \), \( a \), and \( a + d \), where \( a \) is the middle term and \( d \) is the common difference.

2. **Set up the equation for the sum**:
The sum of these numbers is given by:
\[ (a - d) + a + (a + d) = 3a = 24 \]
From this, we can solve for \( a \):
\[ 3a = 24 \implies a = \frac{24}{3} = 8 \]

3. **Set up the equation for the sum of squares**:
The sum of the squares of these numbers is given by:
\[ (a - d)^2 + a^2 + (a + d)^2 = 224 \]
Substituting \( a = 8 \):
\[ (8 - d)^2 + 8^2 + (8 + d)^2 = 224 \]

4. **Expand the squares**:
Expanding the squares:
\[ (8 - d)^2 = 64 - 16d + d^2 \]
\[ 8^2 = 64 \]
\[ (8 + d)^2 = 64 + 16d + d^2 \]
Now, substituting these into the sum of squares:
\[ (64 - 16d + d^2) + 64 + (64 + 16d + d^2) = 224 \]
Simplifying this:
\[ 64 - 16d + d^2 + 64 + 64 + 16d + d^2 = 224 \]
\[ 192 + 2d^2 = 224 \]

5. **Solve for \( d^2 \)**:
Rearranging gives:
\[ 2d^2 = 224 - 192 \]
\[ 2d^2 = 32 \implies d^2 = \frac{32}{2} = 16 \]
Therefore, taking the square root:
\[ d = \pm 4 \]

6. **Find the numbers**:
Now we can find the three numbers for both cases of \( d \):
- If \( d = 4 \):
- The numbers are \( 8 - 4 = 4 \), \( 8 \), and \( 8 + 4 = 12 \).
- If \( d = -4 \):
- The numbers are \( 8 - (-4) = 12 \), \( 8 \), and \( 8 + (-4) = 4 \).

7. **Conclusion**:
The three numbers in A.P. are \( 4, 8, 12 \) or \( 12, 8, 4 \).

Thus, the answer is:
**The numbers are 4, 8, and 12.**