Find four numbers in A.P. whose sum is 28 and the sum of whose squares is 216.

Answer

To find four numbers in Arithmetic Progression (A.P.) whose sum is 28 and the sum of their squares is 216, we can follow these steps:

### Step 1: Define the four numbers in A.P.
Let the four numbers in A.P. be:
- \( a - 3d \)
- \( a - d \)
- \( a + d \)
- \( a + 3d \)

Here, \( a \) is the middle number and \( d \) is the common difference.

### Step 2: Set up the equation for the sum of the numbers.
The sum of these four numbers can be expressed as:
\[ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a \]
According to the problem, this sum equals 28:
\[ 4a = 28 \]

### Step 3: Solve for \( a \).
Dividing both sides by 4 gives:
\[ a = \frac{28}{4} = 7 \]

### Step 4: Set up the equation for the sum of the squares.
Now, we need to find the sum of the squares of these numbers:
\[ (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 \]
Expanding this:
\[ = (a^2 - 6ad + 9d^2) + (a^2 - 2ad + d^2) + (a^2 + 2ad + d^2) + (a^2 + 6ad + 9d^2) \]
Combining like terms:
\[ = 4a^2 + (9d^2 + d^2 + d^2 + 9d^2) + (-6ad - 2ad + 2ad + 6ad) \]
This simplifies to:
\[ = 4a^2 + 12d^2 \]
According to the problem, this sum equals 216:
\[ 4a^2 + 12d^2 = 216 \]

### Step 5: Substitute \( a \) into the equation.
Substituting \( a = 7 \):
\[ 4(7^2) + 12d^2 = 216 \]
Calculating \( 7^2 \):
\[ 4(49) + 12d^2 = 216 \]
This simplifies to:
\[ 196 + 12d^2 = 216 \]

### Step 6: Solve for \( d^2 \).
Subtracting 196 from both sides:
\[ 12d^2 = 216 - 196 \]
\[ 12d^2 = 20 \]
Dividing by 12:
\[ d^2 = \frac{20}{12} = \frac{5}{3} \]

### Step 7: Solve for \( d \).
Taking the square root:
\[ d = \sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3} \]

### Step 8: Find the four numbers.
Now we can find the four numbers:
1. \( a - 3d = 7 - 3\left(\frac{\sqrt{15}}{3}\right) = 7 - \sqrt{15} \)
2. \( a - d = 7 - \frac{\sqrt{15}}{3} \)
3. \( a + d = 7 + \frac{\sqrt{15}}{3} \)
4. \( a + 3d = 7 + 3\left(\frac{\sqrt{15}}{3}\right) = 7 + \sqrt{15} \)

Thus, the four numbers in A.P. are:
\[ 7 - \sqrt{15}, \quad 7 - \frac{\sqrt{15}}{3}, \quad 7 + \frac{\sqrt{15}}{3}, \quad 7 + \sqrt{15} \]