JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Probability

Three numbers are in geometric progression when the square of the middle equals the product of the extremes, so $3N=(N-2)(N+2)=N^2-4$. This reduces to $N^2-3N-4=0$, that is $(N-4)(N+1)=0$. A sum of two dice lies between $2$ and $12$, so the only admissible value is $N=4$. The outcomes giving a sum of $4$ are $(1,3),(2,2),(3,1)$, three of the $36$ equally likely results, so the probability is $\frac{3}{36}=\frac{1}{12}=\frac{4}{48}$. Thus $k=4$.