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The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive? (1) \(n\) is odd. (2) \(a_n\) is positive
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Last edited by Bunuel on 29 Dec 2012, 04:29, edited 1 time in total.
Renamed the topic, edited the question and added OA.



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Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]
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KevinBrink wrote: Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\). The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?We have following sequence: \(a_1=1\); \(a_2=a_1=1\); \(a_3=3\); \(a_4=a_3=3\); \(a_5=5\); \(a_6=a_5=5\); ... Basically we have a sequence of positive and negative odd integers: 1, 1, 3, 3, 5, 5, 7., 7, 9, 9, ... Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(1)+3+(3)=0\). Also notice that odd terms are positive and even terms are negative. (1) \(n\) is odd > as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive > n is odd, so the same as above. Sufficient. Answer: D. Hope it's clear.
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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03 Jan 2013, 22:38
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a1 + (a1) + a3 + (a3) + .......
1. n is odd > last term is an which is positive every other term cancels out 2. an can be +ve only if n is odd which will be the last term same as above hence answer is D



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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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14 Feb 2013, 10:17
maddyboiler wrote: Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value 1,3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive. dont go into complex things. Just visualize the sequence It can be 2 way 1,1, 3,3, 5,5, 7,7 ending in negative term The sum will be zero in this case 1,1, 3,3, 5,5, 7 ending in positive term The sum will be the last term of sequence we have asked is the sum positive ? > is the sequence as per 2nd case ? > is the a(n) odd ? or is the a(n) positive ? both the statements answer these questions so both are sufficient
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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14 Feb 2013, 06:05



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Re: DS GMAT perp sequence [#permalink]
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23 Mar 2009, 22:48
My Answer: D. EACH statement ALONE is sufficient.
Explanation:
As, ak=k if k is odd Hence, a1=1 a3=3, a5=5 etc.
And as ak=ak1 if k is even Hence, a2=a1=1 a4=a3=3 a6=a5=5 etc.
So, if n is even, a1+a2+....+an = a1+(a1)+a3+(a3)+....+an1+(an1) = 0 All terms get canceled.
And if n is odd, a1+a2+....+an = a1+(a1)+a3+(a3)+....+an2+(an2)+an = an As, ak=k if k is odd, an = n if n is odd Only an remains at the end, which is a positive number equal to n.
Hence each statement satisfies it individually.
Now tell me the OA pls.



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Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]
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28 Dec 2012, 07:34
Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards



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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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30 Dec 2012, 05:13
Accountant wrote: The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?
(1) \(n\) is odd. (2) \(a_n\) is positive In a sequence it always helps to observer a first few terms. Given these definitions : a1=1, a2=a1 = 1, a3 = 3, a4=a3 = 3 so it is clear consecutive terms from begining are canceling each other, i.e., 11+22+33 etc Also, the sum is either positive in which case it is equal to the last odd term or it is zero. So knowing either the term is odd or that last term was positive helps us know that sum of the terms are positive D



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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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14 Feb 2013, 05:55
Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value 1,3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive.



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Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]
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20 Nov 2013, 03:29
Bunuel wrote: KevinBrink wrote: Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\). The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?We have following sequence: \(a_1=1\); \(a_2=a_1=1\); \(a_3=3\); \(a_4=a_3=3\); \(a_5=5\); \(a_6=a_5=5\); ... Basically we have a sequence of positive and negative odd integers: 1, 1, 3, 3, 5, 5, 7., 7, 9, 9, ... Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(1)+3+(3)=0\). Also notice that odd terms are positive and even terms are negative. (1) \(n\) is odd > as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive > n is odd, so the same as above. Sufficient. Answer: D. Hope it's clear. Thanks. But could a1 = 1? the question does not state that k>0 so I make a big mistake here... Please help to explain. Thanks a lot!



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Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]
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20 Nov 2013, 03:30
Thanks guys for the explanations of k. That really help me!!



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Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]
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20 Nov 2013, 03:33
Cee0612 wrote: Bunuel wrote: KevinBrink wrote: Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\). The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?We have following sequence: \(a_1=1\); \(a_2=a_1=1\); \(a_3=3\); \(a_4=a_3=3\); \(a_5=5\); \(a_6=a_5=5\); ... Basically we have a sequence of positive and negative odd integers: 1, 1, 3, 3, 5, 5, 7., 7, 9, 9, ... Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(1)+3+(3)=0\). Also notice that odd terms are positive and even terms are negative. (1) \(n\) is odd > as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive > n is odd, so the same as above. Sufficient. Answer: D. Hope it's clear. Thanks. But could a1 = 1? the question does not state that k>0 so I make a big mistake here... Please help to explain. Thanks a lot! We are told that \(a_k=k\) if \(k\) is odd. Now, substitute k=1 and see what you get.
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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24 Dec 2015, 11:24
The sequence is 1, 1, 3, 3, 5, 5.. Statement 1 means the number of terms is odd. From the above, we can see that the odd terms are positives. THeir negative counterparts are the immediate next term, which are even numbered terms. So statement 1 is sufficient for concluding that the sum is positive. Statement 2 suggest the alst term to be positive. In the sequence a positive is followed by the negative of the same value. Sufficient to answer that the sum is positive. hence, D
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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10 May 2017, 17:47
Bunuel wrote: KevinBrink wrote: Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\). The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?We have following sequence: \(a_1=1\); \(a_2=a_1=1\); \(a_3=3\); \(a_4=a_3=3\); \(a_5=5\); \(a_6=a_5=5\); ... Basically we have a sequence of positive and negative odd integers: 1, 1, 3, 3, 5, 5, 7., 7, 9, 9, ... Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(1)+3+(3)=0\). Also notice that odd terms are positive and even terms are negative. (1) \(n\) is odd > as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive > n is odd, so the same as above. Sufficient. Answer: D. Hope it's clear. Hi Bunuel! Regarding Statement 1, if the number of terms in the sequence is odd, is it mandatory to consider the sequence from a1? What I mean is, is to wrong to choose terms at random? For example, if I choose a4, a5 and a6 from the sequence, the sum of the terms is negative. Thank you for all of your help.



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Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]
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10 May 2017, 21:47
nelliegu wrote: Bunuel wrote: KevinBrink wrote: Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=a1=1. Regards Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\). The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=a_{k1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?We have following sequence: \(a_1=1\); \(a_2=a_1=1\); \(a_3=3\); \(a_4=a_3=3\); \(a_5=5\); \(a_6=a_5=5\); ... Basically we have a sequence of positive and negative odd integers: 1, 1, 3, 3, 5, 5, 7., 7, 9, 9, ... Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(1)+3+(3)=0\). Also notice that odd terms are positive and even terms are negative. (1) \(n\) is odd > as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive > n is odd, so the same as above. Sufficient. Answer: D. Hope it's clear. Hi Bunuel! Regarding Statement 1, if the number of terms in the sequence is odd, is it mandatory to consider the sequence from a1? What I mean is, is to wrong to choose terms at random? For example, if I choose a4, a5 and a6 from the sequence, the sum of the terms is negative. Thank you for all of your help. No you cannot do that. We are clearly given that the sequence starts with \(a_1\).
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Re: The sequence a1, a2, a3, ..., an of n integers is such that
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