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Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:00
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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: 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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18 Sep 2012, 10:00
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 \(s_{101}\).
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:05
Answer is E as the sum of any first two terms is 3 & then 6, 6, 12, 12 & so on
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:08
first 8 terms 0,3,3,0,3,3,0,3
sum of every four terms is 0+33+0 = 0 sum of S100 = 0 sum of s101 = 0+3 =3
Answer is 3 (C)



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:10
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Ans is C, This would be a series of 3 , 0, 3,0 ... .
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:12
ANswer C.
S101 = a1+ a2+ .....+ a101
a3=a1a2= 3 a4=0
Follwoing the pattern,,
0,3,3,0,3,3,....
S99 = 0
A100=0
A101=3
Hence,S101=3



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:14
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Answer is C
a = {0,3,3,0,3,3,0,3,3....} repeated every 3 counts s101 = contains 33 triplets + 2 so the sum is 33 * (0+33) + (0+3) = 3
Last edited by jirayr on 18 Sep 2012, 10:35, edited 1 time in total.



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:18
IMo  C pattern 0,3,3, ... sum of 3 terms = 0 sum of 99 terms = 0 100 th = 0 101st = 3 sum = 3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:47
Ans=C, i have considered first upto a12, and observed that s1=0, s2=3, s3=0 and the cycle repeats.. by this we can divide 101 as 99+2 i.e., s99 would be 0 and s100 =0 and s101=3.
Last edited by chetanachethu on 18 Sep 2012, 11:05, edited 2 times in total.



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 10:51
Considering the rule of Cyclicity,we can conclude that the pattern repeat itself after 3 units i.e a1=0,a2=3,a3=3 ; again a4=0,a5=3,a6=3 and so on... Now 101/3 [as the pattern repeat itself after 3 units as above] gibes a quotient 99 and thus sum of a1+a2+..a99=0 [as every 3 units having sum 0]and a100+a101=0+3=3 So, S101=a1+a2+..a99+a100+a101=3. So Answer is 3.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 11:24
Answer : C
We have an = a(n2)  a(n1) a1 = 0 a2 =3
If we use the above formula we get a repetitve series of 3,0,3 as follows a3 = a2 a1 = 30 = 3 a4 = a3  a2 = 33 =0 a5 = a4a3 = 0  3 = 3 a6 = a5  a4 = 3  0 = 30 =3 a7 = a6  a5 = 3  3 = 3 3 = 0 And this trend will continue as follows a8 = 3 , a9 = 3 ....and so on until a101 = 3 If we observe there is a pattern emerging starting a3 .If we sum a particular triplet,say a3 + a4+ a5 = 3+0+3 = 0. Our sum will be 0 Similiarly if we keep adding triplets a6 + a7 + a8 =0 a9 + a10 + a11 = 0 . . . a99 + a100 + a101 = 3 + 0 + 3 = 0. So we can conclude the sum of all the triplets is 0. a3 + .a4 + a5 + .......a101 = 0.
S101 = a1 + a2 + (sum of all the triplets present ) = 0 + 3 + 0 Hence S101 = 3.
Answer C



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 11:38
Bunuel wrote: 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 a3 = a1  a2 a4 = a2  a3 a5 = a3  a4 .. .. .. .. a100 = a98  a99 a101 = a99  a100  Summing together both sides a3+a4+a5+.......+a101 = a1  a100 ( because all other values will be cancelled out in right hand side. For example, a2 from 1st equation will be cancelled out from a2 from 2nd equation, and so on.) adding a1 + a2 both sides a1 + a2 +a3 .............. + a101 = S101 = a1+a2+a1  a100 => 0 + 3 + 0  a100 Now these terms follow a pattern as follows: a3 = 3 a4 = 0 a5 = 3 .. .. .. So, EVEN indexed "a" will be zero. Thus, => 0+3+0+0 = 3 Hence, answer is C.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 11:56
a1=0, a2=3, a3=3, a4=0, a5=3, a6=3....... repetition occurs after every 3rd unit/count. S101=a1+a2+a3....+a99+a100+a101 s101=0+a100+a101 as [101][/3] leaves a remainder 2; therefore sum (a1+a2+a3+...a99)=0 S101=0+0+3 =3, (c)



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 13:08
So I figured out the sequence first a1=0, a2=3, a3=3, a4=0, a5=3, a6=3, a7=0, a8=3, a9=3, a10=0....... Every third term is a 3 for a3, a6, a9, a12, a15, a18, a21, a, 24....... They are all 3. So s102 = 3 therefore s101 = 3 Another is that the sum of the first 10 is 0 s10= s1 + s2 +s3 + s4....... is 0 so every tenth interval is 0, therefore s100=0 The answer is 3 which is option C



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 13:46
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 \(s_{101}\)?[/b]
(A) 3 (B) 0 (C) 3 (D) 201 (E) 303
Answer is C:
Explaination: a(1) = 0 (given) a(2) = 3 (given) a(3) = a(1)  a(2) = 0  3 = 0  3 = 3 a(4) = a(2)  a(3) = 3  3 = 3  3 = 0 a(5) = a(3)  a(4) = 3  0 = 3  0 = 3 a(6) = a(4)  a(5) = 0  3 = 0  3 = 3
So the series is 0, 3, 3, 0, 3, 3, for a(1), a(2), a(3), a(4), a(5) and a(6) respectively. If you look at the series carefully, S(3) = 0 + 3 + (3) = 0 this trend will continue and hence S(3), S(6), S(9)..... S(99) = 0 S(101) = S(99) + a(100) + a(101) = 0 + 0 + 3 = 3 S(101) = 3
OA please?



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 14:06
Answer is C!
The pattern is 0,3,3,0,3,3,0,3,3...
101/3=99 R2
0+3=3



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 18:12
The answer is (C)3
Solution: We have: S101 =a1+ a2+a3+a4…..+a99+a100+a101 = a1+ a2+a1 a2 +a2 a3 +⋯.+a97 a98 +a98 a99 +a99 a100  = a1+ a2+a1 a100  = 3a100  (1)
Then: a1=0; a2=3; a3=a1 a2 =03=3; a4=a2 a3 =33=0; a5=a3 a4 =30=3; Similarly: a4=0; a5=3; a6=a4 a5 =03=3; a7=a_5 a6 =33=0; a8=a6 a7 =30=3; From those results,we can conclude that: a(3n+1)=0;a(3n+2)=3;a(3n)=3
So, a_100=a(3×33+1)=0 (2) From (1) and (2), we have: S101= 3a100  = 3



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 19:29
Answer is C:
Explaination: a(1) = 0 (given) a(2) = 3 (given) a(3) = a(1)  a(2) = 0  3 = 0  3 = 3 a(4) = a(2)  a(3) = 3  3 = 3  3 = 0 a(5) = a(3)  a(4) = 3  0 = 3  0 = 3 a(6) = a(4)  a(5) = 0  3 = 0  3 = 3
However the question asks for S(101). So we have to find a series for S(n). S(1) = a(1) = 0 S(2) = a(1) + a(2) = 0 + 3 = 3 S(3) = a(1) + a(2) + a(3) = 0 + 3  3 = 0 S(4) = a(1) + a(2) + a(3) + a(4) = 0 + 3  3 + 0 = 0 s(5) = a(1) + a(2) + a(3) + a(4) + a(5) = 0 + 3  3 + 0 + 3 = 3 s(6) = a(1) + a(2) + a(3) + a(4) + a(5) +a(6) = 0 + 3  3 + 0 + 3  3 = 0
So the real series is 0, 3, 0, 0, 3, 0, for s(1), s(2), s(3), s(4), s(5) and s(6) respectively.
If you divide 101/3 you get 33 remainder 2, which tells us the correct answer should be the 2nd term of the series, hence S(101) = 3



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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18 Sep 2012, 22:23
Solution
As per the question we have values of terms a1 = 0;a2 =3;a3 = 3
Using the function \(a_{n} = a_{(n2)}  a_{(n1)}\) for all \(n\geq{3}\), can derive
a4 = 0
If we continue solving the next two terms (i.e a5 & a6) we will see that these terms are cyclic i.e (0,3,3)
Moving forward we have,
\(s_{4} = a_1 + a_2 + a_3 + a_4\)
Calculating the first four terms s (4) = 0 ; In the same way s(100) = 0 and s(101) = s(100) + a101 = 3
Hence OA is C



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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]
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19 Sep 2012, 09:01
We have a winner! 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 \(s_{101}\).
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