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(B) The total number of outcomes when two dice are rolled is $n(S) = 6 	imes 6 = 36$.Given that $N - 2, \sqrt{3N}, N + 2$ are in geometric progressio

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$4$

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(B) The total number of outcomes when two dice are rolled is $n(S) = 6 	imes 6 = 36$.Given that $N - 2, \sqrt{3N}, N + 2$ are in geometric progression ($G$.$P$.),the square of the middle term must equal the product of the extremes:$(\sqrt{3N})^2 = (N - 2)(N + 2)$$3N = N^2 - 4$$N^2 - 3N - 4 = 0$Factoring the quadratic equation:$(N - 4)(N + 1) = 0$Since $N$ is the sum of two dice,$N \geq 2$,so $N = 4$ is the only valid solution.The outcomes for which the sum $N = 4$ are: $(1, 3), (3, 1), (2, 2)$.Thus,the number of favorable outcomes is $n(A) = 3$.The probability is $P(A) = \frac{3}{36} = \frac{1}{12}$.We are given $P(A) = \frac{k}{48}$,so:$\frac{k}{48} = \frac{1}{12}$$k = \frac{48}{12} = 4$.

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

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