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by susantaiitk » Mon May 03, 2010 11:03 am
Might be long but this is how I did:

sequence: -1 1 2 1 -1 -2 | -1 1 2 1 -1 -2 | ..... period of 6

so 1000/6 = 166 and rem 4

sum of each period is 0, so sum = sum of last 4 terms i.e -1 1 2 1 = 3
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by iamseer » Mon May 03, 2010 11:06 am
the series is

-1,1,2,1,-1,-2, -1,1,2,1,-1,-2,..... (6 terms adding to 0 repeat 166 times ) and then the last 4 terms -1,1,2,1 remain. They add up to 3.
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by GmatTakerNo.1 » Mon May 03, 2010 11:07 am
Yeah, that´s the way I did it as well, but is there a shorter approach?
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by nakul_anand » Tue May 04, 2010 3:07 am
The first 10 terms of the series are -

(1) -1,1,2,1,-1,-2,-1,1,2,1 and the sum of these 10 terms = 3

(2) The 1st 6 terms -1,1,2,1,-1,-2 repeat every time.

(3) Since 1000 is a multiple of 10, we get 3 as the sum if we calculate A1+A2+....+A1000.
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