In best case there is a a possibility that u can complete ur search at a very first element out of N elements. So Number of comparisons required is O(1).
In worst case u might end up ur search at last element in the list of N elements, so Number of comparisons required is O(N).
So Average Number of comparisons= (O(N) + O(1) ) / 2
= (N+1)/2 (C)