Four different integers form an increasing `A.P` One of these numbers is equal to the sum of the squares of the other three numbers. Then The smallest number is

Options

A. `-2,-1,0,1`
B. 0,1,2,3
C. `-1,0,1,2`
D. none of these

Explanation

To solve the problem, we need to find four different integers that form an increasing arithmetic progression (A.P.) such that one of these integers is equal to the sum of the squares of the other three integers. Let's denote these integers as \( a - d \), \( a \), \( a + d \), and \( a + 2d \), where \( a \) is the middle integer and \( d \) is the common difference.

### Step 1: Set up the equation
According to the problem, one of the integers is equal to the sum of the squares of the other three. We can assume \( a + 2d \) is the integer that equals the sum of the squares of the other three integers:

\[ a + 2d = (a - d)^2 + a^2 + (a + d)^2 \]

### Step 2: Expand the right-hand side
Now, let's expand the right-hand side:

\[ (a - d)^2 = a^2 - 2ad + d^2 \]
\[ a^2 = a^2 \]
\[ (a + d)^2 = a^2 + 2ad + d^2 \]

Adding these together:

\[ (a - d)^2 + a^2 + (a + d)^2 = (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) \]
\[ = 3a^2 + 2d^2 \]

### Step 3: Set up the equation
Now we can set up the equation:

\[ a + 2d = 3a^2 + 2d^2 \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ 3a^2 - a + 2d^2 - 2d = 0 \]

### Step 5: Solve for \( d \)
This is a quadratic equation in terms of \( d \). We can use the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b = -2 \), and \( c = -a + 3a^2 \):

\[ d = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-a + 3a^2)}}{2 \cdot 2} \]
\[ = \frac{2 \pm \sqrt{4 + 8a - 24a^2}}{4} \]
\[ = \frac{2 \pm \sqrt{8a - 24a^2 + 4}}{4} \]

### Step 6: Find integer solutions
For \( d \) to be an integer, the expression under the square root must be a perfect square. We can simplify this further and check for integer values of \( a \).

### Step 7: Testing integer values
Let's test \( a = 0 \):

\[ d = \frac{2 \pm \sqrt{4}}{4} = \frac{2 \pm 2}{4} \]
This gives \( d = 1 \) or \( d = 0 \). Since \( d \) must be greater than 0, we take \( d = 1 \).

### Step 8: Finding the integers
Now we can find the integers:

- \( a - d = 0 - 1 = -1 \)
- \( a = 0 \)
- \( a + d = 0 + 1 = 1 \)
- \( a + 2d = 0 + 2 = 2 \)

Thus, the integers are \( -1, 0, 1, 2 \).

### Conclusion
The smallest number is \( -1 \).