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 $$\frac{k}{48}$$. Then the value of $$k$$ is

Find the value of $$k$$ where the probability that $$N-2, \sqrt{3N}, N+2$$ are in geometric progression is $$\frac{k}{48}$$.

For three terms in GP, the square of the middle term equals the product of the other two: $$(\sqrt{3N})^2 = (N-2)(N+2)$$. This gives $$3N = N^2 - 4$$ and hence $$N^2 - 3N - 4 = 0$$.

The equation factors as $$(N-4)(N+1) = 0$$, giving $$N = 4$$ or $$N = -1$$; since $$N$$ is the sum of two dice, $$N \geq 2$$, we take $$N = 4$$.

For sum 4, the favorable outcomes are $$(1,3), (2,2), (3,1)$$, yielding 3 outcomes out of a total of $$6 \times 6 = 36$$. Hence $$P(N=4) = \frac{3}{36} = \frac{1}{12} = \frac{4}{48}$$.

Equating $$\frac{k}{48} = \frac{4}{48}$$ gives $$k = 4$$, so the correct answer is Option B: $$4$$.