Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that N-2,√3N,N+2 are in geometric progression be k/48. Then the value of k is :

Solution

Maths sir · 4 steps

Use the GP condition, middle term squared equals the product of the ends, to find the one allowed N.

1. GP condition. Three terms a, b, c are in GP when b² = ac. Here that means (√3N)² = (N-2)(N+2).

   3N = N² - 4 ⇒ N² - 3N - 4 = 0 ⇒ (N-4)(N+1) = 0

2. Pick valid N. The sum of two dice lies between 2 and 12, so N = -1 is impossible and N = 4. Check: the terms are 2, √12, 6, and the common ratio is √3 both times.
3. Count outcomes. A sum of 4 comes from (1,3), (2,2) and (3,1), which gives 3 of the 36 equally likely outcomes. So the probability is 3/36 = 1/12 = 4/48.
4. So. k = 4, which is option (d).

The idea to keep. Turn a GP condition into b² = ac, solve, then keep only the values the dice can produce. With two dice, count ordered pairs out of 36.

Answer from the NTA final answer key. Solution by PrepWiser.