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Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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18 Mar 2007, 16:41

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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?

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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18 Mar 2007, 17:00

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(D) for me

Let us describe the few first terms to have a better idea of it works:
o a(1) = 1
o a(2) = -1
o a(3) = 3
o a(4) = -3

So,
o If n is even, then the sum of a(k) terms give 0. We always have couples of opposite number in the sequence.
o If n is odd, then the sum will be equal to n. All other numbers are in couple (negative/positive), giving 0 if we add them.

From 1 n is odd. Bingo, the sum is positive.

SUFF.

From 2 a(n) > 0... Then n must be an odd. Bingo, the sum is positive.

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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31 Aug 2007, 12:51

jp888 wrote:

The sequence a1, a2, a3,...an of n integers is such that ak = k if k is odd, and ak = -ak-1 if k is even. Is the sum of the terms in the sequence positive?

1) n is odd 2) an is positive

*note, figures after 'a' are in subscript, e.g. a1 and ak-1

this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + .....

so if n is odd all will cancel except the last positive number.
So 1) is sufficient

if an is positive, that is the only one remaining because the rest all pairs cancel out. So 2) is sufficent too...

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 May 2011, 11:47

carpeD wrote:

jp888 wrote:

The sequence a1, a2, a3,...an of n integers is such that ak = k if k is odd, and ak = -ak-1 if k is even. Is the sum of the terms in the sequence positive?

1) n is odd 2) an is positive

*note, figures after 'a' are in subscript, e.g. a1 and ak-1

this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + .....

so if n is odd all will cancel except the last positive number. So 1) is sufficient

if an is positive, that is the only one remaining because the rest all pairs cancel out. So 2) is sufficent too...

Both are individually sufficient

Hi, Could you please explain to me the decomposition part " this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + ....."

If ak=k when odd, then a1= 1 and ak=-ak-1 when even then a2 should be -2-1 =-3 . Therefore the sequence should be 1, -3, 3, -5, 5.... ?? not sure what im missing. please help. Thanks

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 May 2011, 12:24

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I'll take a shot at explaining ...

If k is odd we know all the values are +ve and are equal to k (e.g. 1,3,5,7...) If k is even we know all the values are -ve and are equal to the value of the prior term (e.g. a2 = -1,a4=-3... so the values will be, -1,-3,-5,-7 ....)

1, -1, 3, -3, 5, -5 .....

So as you can see at this point we know that for every value of k (when odd) we have a -ve value from when K is even, unless N (total terms) is odd in which case we will have one extra +ve term that will not cancel out. So if we have even number terms we know the result will be 0 (which is not positive). Therefore to get a positive sum we need one extra odd term.

Try it out,

N=5 1, -1, 3, -3, 5 (if add them, everything cancels out except 5, which is positive).

N=6 1, -1, 3, -3, 5, -5 (if add them, everything cancels out, result is not positive).

So, before looking at the statements we are able to rephrase the question to: "Is the number terms in the sequence odd?"

Statement 1: Gives us exactly that, therefore sufficient. Statement 2: Well, it gives the same thing but instead of saying the number of terms is odd, it says the last term is +ve, which means the same thing as per our sequence above, there sufficient.

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 May 2011, 13:23

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Carol680 wrote:

Hi, Could you please explain to me the decomposition part " this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + ....."

If ak=k when odd, then a1= 1 and ak=-ak-1 when even then a2 should be -2-1 =-3 . Therefore the sequence should be 1, -3, 3, -5, 5.... ?? not sure what im missing. please help. Thanks

\(A_{1}=1\) \(A_{2}=-A_{1}=-1\)

\(A_{3}=3\) \(A_{4}=-A_{3}=-3\)

\(A_{5}=5\) \(A_{6}=-A_{5}=-5\)

\(A_{7}=7\) \(A_{8}=-A_{7}=-7\) . . .

What do we see here: \(A_1+A_2=1-1=0\) \(A_3+A_4=3-3=0\) \(A_5+A_6=5-5=0\) ...

Thus, if we have even number of elements in the series, their addition will always result in 0. If we have odd number of elements, their addition will always result in +ve.

Q: Is there odd number of elements?

1. n is odd. Precisely what we wanted to know. Sufficient.

2. \(A_n\) is positive. Means, the last element in the series is +ve. We know, only odd number(index) has +ve values. All even values have -ve value. Thus, there are odd number of elements. Sufficient.

Ans: "D" **********************

By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero. **********************************
_________________

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 Jul 2011, 03:01

fluke wrote:

Carol680 wrote:

Hi, Could you please explain to me the decomposition part " this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + ....."

If ak=k when odd, then a1= 1 and ak=-ak-1 when even then a2 should be -2-1 =-3 . Therefore the sequence should be 1, -3, 3, -5, 5.... ?? not sure what im missing. please help. Thanks

\(A_{1}=1\) \(A_{2}=-A_{1}=-1\)

\(A_{3}=3\) \(A_{4}=-A_{3}=-3\)

\(A_{5}=5\) \(A_{6}=-A_{5}=-5\)

\(A_{7}=7\) \(A_{8}=-A_{7}=-7\) . . .

What do we see here: \(A_1+A_2=1-1=0\) \(A_3+A_4=3-3=0\) \(A_5+A_6=5-5=0\) ...

Thus, if we have even number of elements in the series, their addition will always result in 0. If we have odd number of elements, their addition will always result in +ve.

Q: Is there odd number of elements?

1. n is odd. Precisely what we wanted to know. Sufficient.

2. \(A_n\) is positive. Means, the last element in the series is +ve. We know, only odd number(index) has +ve values. All even values have -ve value. Thus, there are odd number of elements. Sufficient.

Ans: "D" **********************

By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero. **********************************

Hi Fluke,

How will the ans be D here ? By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero.

If n is even then all the terms cancel out so the sum of terms in the sequence is neither positive nor negative...so I is insufficient ..correct?? or its sufficient since we can definitely answer yes or no ???

What about statement 2 : If an is -ve then the sum of terms is also 0 here so same as case I ....it should be insufficient ...correct??

or the logic here is that since we can definitely answer both the statements its D.... PLease let me know...

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 Jul 2011, 03:52

siddhans wrote:

fluke wrote:

Carol680 wrote:

Hi, Could you please explain to me the decomposition part " this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + ....."

If ak=k when odd, then a1= 1 and ak=-ak-1 when even then a2 should be -2-1 =-3 . Therefore the sequence should be 1, -3, 3, -5, 5.... ?? not sure what im missing. please help. Thanks

\(A_{1}=1\) \(A_{2}=-A_{1}=-1\)

\(A_{3}=3\) \(A_{4}=-A_{3}=-3\)

\(A_{5}=5\) \(A_{6}=-A_{5}=-5\)

\(A_{7}=7\) \(A_{8}=-A_{7}=-7\) . . .

What do we see here: \(A_1+A_2=1-1=0\) \(A_3+A_4=3-3=0\) \(A_5+A_6=5-5=0\) ...

Thus, if we have even number of elements in the series, their addition will always result in 0. If we have odd number of elements, their addition will always result in +ve.

Q: Is there odd number of elements?

1. n is odd. Precisely what we wanted to know. Sufficient.

2. \(A_n\) is positive. Means, the last element in the series is +ve. We know, only odd number(index) has +ve values. All even values have -ve value. Thus, there are odd number of elements. Sufficient.

Ans: "D" **********************

By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero. **********************************

Hi Fluke,

How will the ans be D here ? By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero.

If n is even then all the terms cancel out so the sum of terms in the sequence is neither positive nor negative...so I is insufficient ..correct?? or its sufficient since we can definitely answer yes or no ???

What about statement 2 : If an is -ve then the sum of terms is also 0 here so same as case I ....it should be insufficient ...correct??

or the logic here is that since we can definitely answer both the statements its D.... PLease let me know...

Actually my bad,

2. \(A_n\) is \(0\) {Note: \(A_n\) can't be negative.} Q: Is there odd number of elements? A: No. Because Sum=0; number of elements must be even. Sufficient.

1. n is even. Q: Is there odd number of elements? A: No. We are given the answer here. Sufficient.
_________________

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: \(a_1=1\); \(a_2=-a_1=-1\); \(a_3=3\); \(a_4=-a_3=-3\); \(a_5=5\); \(a_6=-a_5=-5\); ...

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: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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19 May 2012, 02:45

The way the sequence has been defined, a(k) + a(k+1) will always be 0 for every odd integer k.

Stt 1: If n is odd, that means the last term in the series is odd. As the sum of all preceding terms has to be zero, and a(k) is always positive when k is odd, the sum is always positive. Sufficient.

Stt 2: If a(n) is +ve, this means n is odd. By the same logic, the sum is always positive. Sufficient.

D it is.

@agdimple33: Yes, it should be n consecutive integers.
_________________

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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29 Jan 2017, 03:47

(1) n is odd If n is 5, for instance, then a5=5, a4=-3, a3=3, a2=-1, a1=1 (I used each of the equations listed above to get these numbers I just followed the pattern) When added together [5+(-3)+(3)+(-1)+(1)] the result is +ve5 You could also plug-in 7 for n to make sure and you would still get a +ve result-follow the same pattern. This statement alone is sufficient.

(2) an is +ve/ means the result must be +ve not zero or -ve

If n=5 (you get the same result as in statement 1) However, if you let n=4/ following the same procedure as above (statement 1) you will see that the result is 0; therefore, one can conclude that n must be an odd number to get a +ve result. n cannot be even. This statement alone is also sufficient.

Concentration: General Management, International Business

GMAT 1: 650 Q47 V32

GPA: 3.87

Re: Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]

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23 Apr 2017, 14:53

fluke wrote:

Carol680 wrote:

Hi, Could you please explain to me the decomposition part " this decomposes to 1 + (-1) + 3 + (-3) + 5 + (-5) + ....."

If ak=k when odd, then a1= 1 and ak=-ak-1 when even then a2 should be -2-1 =-3 . Therefore the sequence should be 1, -3, 3, -5, 5.... ?? not sure what im missing. please help. Thanks

\(A_{1}=1\) \(A_{2}=-A_{1}=-1\)

\(A_{3}=3\) \(A_{4}=-A_{3}=-3\)

\(A_{5}=5\) \(A_{6}=-A_{5}=-5\)

\(A_{7}=7\) \(A_{8}=-A_{7}=-7\) . . .

What do we see here: \(A_1+A_2=1-1=0\) \(A_3+A_4=3-3=0\) \(A_5+A_6=5-5=0\) ...

Thus, if we have even number of elements in the series, their addition will always result in 0. If we have odd number of elements, their addition will always result in +ve.

Q: Is there odd number of elements?

1. n is odd. Precisely what we wanted to know. Sufficient.

2. \(A_n\) is positive. Means, the last element in the series is +ve. We know, only odd number(index) has +ve values. All even values have -ve value. Thus, there are odd number of elements. Sufficient.

Ans: "D" **********************

By the way, even if the statement said; 1. n is even 2. \(A_n\) is -ve.

The answer would be "D" because we would definitely know that the sum of terms is not +ve. It's zero. **********************************

]]

Hello, Fluke (and everyone else).

Can you guys help me out with a question about this question? I understand the rationale used, except for one thing: Could we conclude that the fist statement (N is odd) be an odd negative number? > For example: n=-5