Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Sequence a1, a2, a3....an of n integers is such that ak = k [#permalink]
18 Mar 2007, 15:41

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

52% (02:04) correct
48% (01:16) wrong based on 112 sessions

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?

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: Sequences - GMATPrep [#permalink]
31 Aug 2007, 11:51

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: Sequences - GMATPrep [#permalink]
29 May 2011, 10:47

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: Sequences - GMATPrep [#permalink]
29 May 2011, 11:24

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: Sequences - GMATPrep [#permalink]
29 May 2011, 12:23

1

This post received KUDOS

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, 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...

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. _________________

Re: Sequences - GMATPrep [#permalink]
18 May 2012, 01:48

3

This post received KUDOS

Expert's post

mila84 wrote:

why do everyone assume that a1 is positive?

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]
19 May 2012, 01:45

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]
26 Apr 2014, 07:12

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________