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# Veritas Prep 10 Year Anniversary Promo Question #3

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Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]  18 Sep 2012, 09:00
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Veritas Prep 10 Year Anniversary Promo Question #3

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course (\$1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

A sequence is given by the rule $$a_{n} = |a_{(n-2)}| - |a_{(n-1)}|$$ for all $$n\geq{3}$$, where $$a_1 = 0$$ and $$a_2 = 3$$.

A function $$s_{n}$$ is defined as the sum of all the terms of the sequence from its beginning through $$a_n$$. For instance, $$s_{4} = a_1 + a_2 + a_3 + a_4$$. What is $$s_{101}$$?

(A) -3
(B) 0
(C) 3
(D) 201
(E) 303
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]  20 Sep 2012, 00:01
Because the series is ~ 0,3,-3,0,3,-3,0 .... Upto infinite .. Therefore we can imagine it as a set where the sum of every third is ZERO , so after every 3 terms we start from scratch with a zero... At A99 , the sum will be zero , and the 100th term will be a 0 , followed by the 101st term ie a 3 therfore from numbers 1 thru 99 the sum = 0 , from 100-101 the sum is 0+3 ie. 3. Therefore the answer is +3 ...

We an also devide the closest term to 101 by 3 , ie 99 so start with from a clean state from 99 , and just calculate the sum of 100 and 101st term... So the sum of the first two terms of the series ... That is S2 .. which is 0 +3 = 3 ...
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]  04 Jan 2015, 17:03
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Re: Veritas Prep 10 Year Anniversary Promo Question #3   [#permalink] 04 Jan 2015, 17:03

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