Question

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

• A. 16
• B. 2
• C. 8
• D. 4

Answer 1

The Correct Option is D

To solve this problem, we start by understanding what it means for the numbers \(N-2\), \(\sqrt{3N}\), and \(N+2\) to be in geometric progression. In a geometric progression, each term after the first is the product of the previous term and a constant ratio. Let's denote this common ratio by \(r\).

So, for the sequence \(N-2\), \(\sqrt{3N}\), \(N+2\), we have:

1. \(\sqrt{3N} = r(N-2)\)
2. \(N+2 = r \sqrt{3N}\)

Solving these two equations will give us the value of \(N\).

Step 1: Solve the First Equation

\(\sqrt{3N} = r(N-2)\)

Square both sides to eliminate the square root:

\(3N = r^2(N-2)^2\).

Step 2: Solve the Second Equation

\(N+2 = r \sqrt{3N}\)

Substitute \(\sqrt{3N} = r(N-2)\) from Step 1 into this equation:

\(N + 2 = r \cdot r(N - 2) = r^2(N - 2)\).

Step 3: Express \(r^2\)

From both the equations after simplification:

• \(3N = r^2(N-2)^2\)
• \(N + 2 = r^2(N-2)\)

Equating for \(r^2\) from the second equation:

\(r^2 = \frac{N+2}{N-2}\).

Substitute back to find N

From equation one, let's replace \(r^2\) in \(3N = r^2(N-2)^2\):

\(3N = \left(\frac{N+2}{N-2}\right) (N-2)^2\)

Solving:

\(3N = (N+2)(N-2)

\(3N = N^2 - 4\)

Rearrange terms:

\(N^2 - 3N - 4 = 0\)

Step 4: Solve Quadratic Equation

Factor the quadratic \(N^2 - 3N - 4 = 0\):

\((N - 4)(N + 1) = 0\)

The solutions give \(N = 4\) and \(N = -1\). Only \(N = 4\) is viable since \(N\) must be a positive number representing the sum of the dice.

Step 5: Determine Probability

The possible sums \(N\) of two dice are 2 through 12. Our valid \(N = 4\).

Sum \(N\)	Combinations
2	(1,1)
3	(1,2), (2,1)
4	(1,3), (2,2), (3,1)

Since there are 3 combinations to get a sum of 4, and each combination's occurrence in dice rolling is equally likely with a total of 36 outcomes. Probability is:

\(\frac{3}{36} = \frac{1}{12}\).

Conclusion 

We set this probability equation equals to \(\frac{k}{48}\) as given:

\(\frac{1}{12} = \frac{k}{48}\).

Solving for \(k\):

\(k = \frac{48}{12} = 4\).

Thus, the value of \(k\) is 4.