a, b, c, d and e are 5 distinct numbers that from an arithmetic progression.…

Let the five terms of the arithmetic progression be T1, T2, T3, T4, T5 where Tn = p + (n-1)q.

Because c is the arithmetic mean of a and b, the index of c equals the average of the indices of a and b. Because d is the arithmetic mean of b and c, the index of d equals the average of the indices of b and c. Solving these integer index constraints with indices 1..5 forces c to be the middle term T3.

Therefore T1, T2, T3, T4, T5 are the five values and c = T3.

Statement i: The average of all five terms is (T1+T2+T3+T4+T5)/5 = T3, because in any odd-length consecutive AP the mean equals the middle term. Hence the average of all five terms is c. Statement i is true.

Statement ii: Evaluate the two averages using the positions forced by the mean relations. In the two possible arrangements compatible with the mean conditions the order is either b, d, c, e, a or a, e, c, d, b. In either arrangement (d+e)/2 = (T2+T4)/2 = T3 and (a+b)/2 = (T5+T1)/2 = T3, so (d+e)/2 = (a+b)/2 and therefore (d+e)/2 is not greater than (a+b)/2. Statement ii is true.

Statement iii: Compare (b+c)/2 and (a+d)/2. Using the arrangement b = T1, c = T3, a = T5, d = T2 (one valid arrangement), we get (b+c)/2 = (T1+T3)/2 = p+q and (a+d)/2 = (T5+T2)/2 = p+2.5q. The inequality (b+c)/2 > (a+d)/2 would require q < 0 in this arrangement; for q > 0 it fails. Thus statement iii depends on the sign of the common difference and is not always true.

A concrete counterexample: take the AP 0,1,2,3,4 and the arrangement b=0, d=1, c=2, e=3, a=4. Then (b+c)/2 = 1 while (a+d)/2 = 2.5, so (b+c)/2 is not greater than (a+d)/2. Therefore statement iii is false in general.

Conclusion: Only statements i and ii are always true. The correct choice is the one that states i and ii only.