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by diehard_gmat » Thu Feb 03, 2011 7:57 am
Consider the sequence 1, -2, 3, -4, 5, -6, ..... , whose n-th term is [(-1)^(n+1)]*(n). What is the average of the first 200 terms of the sequence?

A) -1
B) -0.5
C) 0
D) 0.5
E) 1
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by Anurag@Gurome » Thu Feb 03, 2011 7:59 am
diehard_gmat wrote:Consider the sequence 1, -2, 3, -4, 5, -6, ..... , whose n-th term is [(-1)^(n+1)]*(n). What is the average of the first 200 terms of the sequence?

A) -1
B) -0.5
C) 0
D) 0.5
E) 1
Note that the sum of alternate pairs of the series taken in order is -1.
There are 100 such pairs in the first 200 terms. Sum of each pair is -1.
Hence, sum of 200 terms = (-1)*(100) = -100

Hence, average = (-100)/200 = -0.5

The correct answer is B
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by aleph777 » Fri Feb 04, 2011 6:25 am
Maybe not as technical as Anurag's explanation, but since I knew this was an average problem, I knew I could find the average using the formula largest number + smallest number / 2. After plugging in 200 into the formula, though, and pulling out -200, I realized that would have to be the lowest. I looked back at the original pattern, and noticed every odd digit was positive, and every even digit was negative, so if -200 is the SMALLEST number, then by plugging in 199, you'd find the largest digit (which is just 199, after computation).

So then I simply plugged these two into the standard average formula:

[199 + (-200)]/2 = -0.5
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