Let `N` be the sum of the numbers appeared when two fair dice are rolled and let the probability that `N-2,sqrt(3N),N+2` are in geometric progression be `(k)/(48)`.Then the value of `k` is

Options

A. `16`
B. `2`
C. `8`
D. `4`

Answer

To solve the problem, we need to find the value of \( k \) such that the probability that \( N-2, \sqrt{3N}, N+2 \) are in geometric progression is \( \frac{k}{48} \).

### Step 1: Understanding the Sum \( N \)
When two fair dice are rolled, the possible sums \( N \) can range from 2 (1+1) to 12 (6+6). Thus, \( N \) can take values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

### Step 2: Setting Up the Condition for Geometric Progression
For three numbers \( a, b, c \) to be in geometric progression, the condition is:
\[ b^2 = ac \]
In our case, we have:
- \( a = N - 2 \)
- \( b = \sqrt{3N} \)
- \( c = N + 2 \)

Thus, we need to check:
\[ (\sqrt{3N})^2 = (N - 2)(N + 2) \]
This simplifies to:
\[ 3N = N^2 - 4 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ N^2 - 3N - 4 = 0 \]

### Step 4: Solving the Quadratic Equation
Now we can solve the quadratic equation using the quadratic formula:
\[ N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -3, c = -4 \):
\[ N = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ N = \frac{3 \pm \sqrt{9 + 16}}{2} \]
\[ N = \frac{3 \pm \sqrt{25}}{2} \]
\[ N = \frac{3 \pm 5}{2} \]
Thus, we have two potential solutions:
\[ N = \frac{8}{2} = 4 \quad \text{and} \quad N = \frac{-2}{2} = -1 \]

### Step 5: Validating the Solutions
Since \( N \) must be a positive integer (the sum of two dice), we discard \( N = -1 \) and keep \( N = 4 \).

### Step 6: Finding the Probability
Now, we need to find the probability that \( N = 4 \) when rolling two dice. The possible combinations that yield a sum of 4 are:
- (1, 3)
- (2, 2)
- (3, 1)

There are 3 combinations that yield a sum of 4. The total number of outcomes when rolling two dice is \( 6 \times 6 = 36 \). Therefore, the probability \( P(N = 4) \) is:
\[ P(N = 4) = \frac{3}{36} = \frac{1}{12} \]

### Step 7: Relating Probability to \( k \)
We are given that this probability can be expressed as \( \frac{k}{48} \). Setting these equal gives:
\[ \frac{1}{12} = \frac{k}{48} \]
Cross-multiplying yields:
\[ k = \frac{48}{12} = 4 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{4} \]