Data Sufficiency

For a certain set of n numbers, where n > 2, is the average
(arithmetic mean) equal to the median?
(1) The n numbers are positive, consecutive even
integers.
(2) The average of the n numbers is equal to the
average of the largest and smallest numbers
in the set.

[spoiler]OA:A[/spoiler]
[spoiler]
why is statement 2 insufficient?
i tried different combinations of numbers and got mean=median:
e.g. 3,5,5,7 median=5, mean=5 mean=median . even though the set is not evenly spaced
3,4,6,7 median=5, mean=5 mean=median . also,set is not evenly spaced
i don't find a case to disprove statement 2[/spoiler]

An alternate way to evaluate statement 2:

(2) The average of the n numbers is equal to the
average of the largest and smallest numbers
in the set.

Let n=5, implying a set of 5 numbers.
Let the 5 numbers, in ascending order, be a, b, c, d and e.

Since the average of the 5 numbers is equal to the average of a and e, we get:
(a+b+c+d+e)/5 = (a+e)/2
2(a+b+c+d+e) = 5(a+e)
2(b+c+d) + 2(a+e) = 5(a+e)
2(b+c+d) = 3(a+e)
(a+e)/(b+c+d) = 2/3.

Let a+e = 8 and b+c+d = 12, so that (a+e)/(b+c+d) = 8/12 = 2/3.
Average of the 5 numbers = (a+b+c+d+e)/5 = (8+12)/5 = 4.
Let a=0 and e=8, so that a+e = 8.

Case 1: The numbers are evenly spaced
Let b=2, c=4, and d=6, so that b+c+d = 12.
Resulting set:
0, 2, 4, 6, 8.
In this case, the average of the 5 numbers (4) is equal to the median (also 4).

Case 2: The numbers are NOT evenly spaced
Let b=3, c=3, and d=6, so that b+c+d = 3+3+6 = 12.
Resulting set:
0, 3, 3, 6, 8.
In this case, the average of the 5 numbers (4) is NOT equal to the median (3).

Thus, statement 2 is INSUFFICIENT.