Given,
Two fair dices are rolled,
So total sample space will be n(s)=36
Now given: N−2,3N,N+2 are in G.P.
So, 3N=(N−2)(N+2)
⇒3N=N2−4
⇒N2−3N−4=0
⇒(N−4)(N+1)=0
⇒N=4 or N=−1 (rejected)
Now favourable cases for sum 4≡(1,3),(3,1),(2,2)
So, n(A)=3
Hence, P(A)=363=121=484⇒k=4