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The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

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23 Mar 2009, 11:17

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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 \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

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 \(a_k=-a_{k-1}\) 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.

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

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
_________________

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.

\(k\) in \(a_k\) is a subscript, meaning that \(a_k\) is \(k_{th}\) term in the given sequence which starts from \(a_1\), thus k must be some positive integer.

And as ak=-ak-1 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)+....+an-1+(-an-1) = 0 All terms get canceled.

And if n is odd, a1+a2+....+an = a1+(-a1)+a3+(-a3)+....+an-2+(-an-2)+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.

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 \(a_k=-a_{k-1}\) 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., 1-1+2-2+3-3 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

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.

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 \(a_k=-a_{k-1}\) 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!

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 \(a_k=-a_{k-1}\) 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.
_________________

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

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10 Dec 2014, 01:05

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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.

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

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27 Feb 2017, 20:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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 \(a_k=-a_{k-1}\) 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.

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 \(a_k=-a_{k-1}\) 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 \(a_1\).
_________________