successful linear search complexity

Question

Average number of comparisons required for successful linear search in table of length N
a.( N+2)/2
b. N(N+1)/2
c. N-1/2
d. None of these

Comments

in best case,we will have 1 comparison, and in worst case we will have 'n' comparisons.so average = n+1/2

Answer 1

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)