gmatophobia
Nov 07 - Question of the day The median of n consecutive odd integers is 30. If the fifth term is 33, then which of the following can be the last term of the sequence?
A.19
B.21
C.25
D.33
E.41
Source: eGMAT | Difficulty : Hard
Well, the trick here is realizing that it is a descending order AP with consecutive odd integers
Median of 2 odd integers = 30 => That number of terms is EVEN (because if it were odd, then median would have been the middle term and thus odd as well) = (29+31)/2
And initially you think that Since a5 = 33, so 31 = a4 and 29 = a3
Since median of even set of numbers = Average of (n/2) and (n/2 + 1)th terms, WHICH WOULD GIVE US n = 6, and a6 would be 35 BUT THERE IS NO OPTION LIKE THAT and most options are below 33
So that means that a5 = 33, and 31 = a6 and 29 = a7, So n = 12
Calculate a12 would give us 19
Answer: A