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GRE 23- (The average of five consecutive integers starting from m) – (the average of six consecutive integers

By spartan
(The average of five consecutive integers starting from m) – (the average of six consecutive integers
starting from m) =
(A) –1/4
(B) –1/2
(C) 0
(D) 1/2
(E) 1/4
Choose any five consecutive integers, say, –2, –1, 0, 1 and 2. (We chose these particular numbers to
make the calculation as easy as possible. But any five consecutive integers will do. For example, 1, 2, 3, 4,
and 5.) Forming the average yields (–1 + (–2) + 0 + 1 + 2)/5 = 0/5 = 0. Now, add 3 to the set to form 6
consecutive integers: –2, –1, 0, 1, 2, and 3. Forming the average yields
(−1+ (−2) + 0 +1+ 2 + 3)/6=([−1+ (−2) + 0 +1+ 2]+ 3)/6=([0]+ 3)/6= 
since the average of –1 + (–2) + 0 + 1 + 2 is zero, their sum must be zero
3/6 =1/2
(The average of five consecutive integers starting from m) – (The average of six consecutive integers
starting from m) = (0) – (1/2) = –1/2.
The answer is (B).
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