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Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:00
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: veritasprep10yearanniversarygiveaway138806.htmlTo 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 \(a_{n} = a_{(n2)}  a_{(n1)}\) 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]
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20 Sep 2012, 01:01
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 100101 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]
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26 Aug 2016, 04:22



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Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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11 Oct 2017, 13:43
Bunuel wrote: Winner: yogeshwar007Official Explanation: Answer is CWhile we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “preprevious” one. So \(a_3 = a_1  a_2\), and \(a_4 = a_2  a_3\), and \(a_5 = a_3  a_4\), and so on. We are told that the sequence begins 0, 3, ..., so we can derive that: \(a_3 = a_1  a_2 = 0  3 = 0  3 = 3\). Then \(a_4 = a_2  a_3 = 3  3 = 3  3 = 0\). Then \(a_5 = a_3  a_4 = 3  0 = 3  0 = 3\). As soon as we’ve seen \(a_4\) and \(a_5\) turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as \(a_1\) and \(a_2\)) — so \(a_6\) will be identical to \(a_3\), so 3. So we have established the pattern 0, 3, 3, 0, 3, 3, ... for our sequence. Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + 3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with \(a_{99}\). \(a_{100}\) will then add itself (0) to that sum, and \(a_{101}\) will add itself (3) onto that. So we will land at a sum of 3 for \(s_{101}\). hi Bunuel The sum of numbers until a99 will be zero, that's okay, but how can a100 be zero..? At a99 the cycle is complete, the new cycle begins at a100, then how ..? please say to me I have tried this way 99 is a multiple of 3, so when the cycle is complete a99 corresponds to a3, and 100 is multiple of 4, so 100 corresponds to a4 ending at 0, so a101 is equal to 3 please help me work it out thanks in advance, man



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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11 Oct 2017, 20:57
gmatcracker2017 wrote: Bunuel wrote: Winner: yogeshwar007Official Explanation: Answer is CWhile we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “preprevious” one. So \(a_3 = a_1  a_2\), and \(a_4 = a_2  a_3\), and \(a_5 = a_3  a_4\), and so on. We are told that the sequence begins 0, 3, ..., so we can derive that: \(a_3 = a_1  a_2 = 0  3 = 0  3 = 3\). Then \(a_4 = a_2  a_3 = 3  3 = 3  3 = 0\). Then \(a_5 = a_3  a_4 = 3  0 = 3  0 = 3\). As soon as we’ve seen \(a_4\) and \(a_5\) turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as \(a_1\) and \(a_2\)) — so \(a_6\) will be identical to \(a_3\), so 3. So we have established the pattern 0, 3, 3, 0, 3, 3, ... for our sequence. Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + 3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with \(a_{99}\). \(a_{100}\) will then add itself (0) to that sum, and \(a_{101}\) will add itself (3) onto that. So we will land at a sum of 3 for \(s_{101}\). hi Bunuel The sum of numbers until a99 will be zero, that's okay, but how can a100 be zero..? At a99 the cycle is complete, the new cycle begins at a100, then how ..? please say to me I have tried this way 99 is a multiple of 3, so when the cycle is complete a99 corresponds to a3, and 100 is multiple of 4, so 100 corresponds to a4 ending at 0, so a101 is equal to 3 please help me work it out thanks in advance, man It's very easy. The sequence is: {0, 3, 3} {0, 3, 3} {0, 3, 3} {0, 3, 3} ... Notice that if n is a multiple of 3, so a3, a6, ..., a99, then it's 3. The next terms, so a4, a7, ..., a100 are 0.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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12 Oct 2017, 10:04
Bunuel wrote: gmatcracker2017 wrote: Bunuel wrote: Winner: yogeshwar007Official Explanation: Answer is CWhile we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “preprevious” one. So \(a_3 = a_1  a_2\), and \(a_4 = a_2  a_3\), and \(a_5 = a_3  a_4\), and so on. We are told that the sequence begins 0, 3, ..., so we can derive that: \(a_3 = a_1  a_2 = 0  3 = 0  3 = 3\). Then \(a_4 = a_2  a_3 = 3  3 = 3  3 = 0\). Then \(a_5 = a_3  a_4 = 3  0 = 3  0 = 3\). As soon as we’ve seen \(a_4\) and \(a_5\) turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as \(a_1\) and \(a_2\)) — so \(a_6\) will be identical to \(a_3\), so 3. So we have established the pattern 0, 3, 3, 0, 3, 3, ... for our sequence. Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + 3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with \(a_{99}\). \(a_{100}\) will then add itself (0) to that sum, and \(a_{101}\) will add itself (3) onto that. So we will land at a sum of 3 for \(s_{101}\). hi Bunuel The sum of numbers until a99 will be zero, that's okay, but how can a100 be zero..? At a99 the cycle is complete, the new cycle begins at a100, then how ..? please say to me I have tried this way 99 is a multiple of 3, so when the cycle is complete a99 corresponds to a3, and 100 is multiple of 4, so 100 corresponds to a4 ending at 0, so a101 is equal to 3 please help me work it out thanks in advance, man It's very easy. The sequence is: {0, 3, 3} {0, 3, 3} {0, 3, 3} {0, 3, 3} ... Notice that if n is a multiple of 3, so a3, a6, ..., a99, then it's 3. The next terms, so a4, a7, ..., a100 are 0. thanks Bunuel if started from "0", a99 will correspond to "3" and will sum to "0", then a100 will have value "0" and 101 will be "3" now it is very clear to me, thank to you again, man can you please, however, provide me some questions of this kind for further practice? thanks



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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12 Oct 2017, 10:06



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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12 Oct 2017, 10:18
Bunuel wrote: gmatcracker2017 wrote: thanks Bunuel if started from "0", a99 will correspond to "3" and will sum to "0", then a100 will have value "0" and 101 will be "3" now it is very clear to me, thank to you again, man can you please, however, provide me some questions of this kind for further practice? thanks 12. Sequences thanks a lot bunu




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