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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

ans is A: REASON let S=sum of 'n' consecutive nos with 'a' as first no = n(2a+(n-1)d)/2... so S/n= n(2a+(n-1)d)/2n=(2a+(n-1)d)/2, which is equal to the avg of n nos....

now in consecutive nos avg is the center no if n is odd or avg of center two nos,which would be in decimals((odd + even)/2) if n is even.. by statement I...avg of n consecutive nos is an odd no... therefore n is odd .. so sum is odd no *odd no= odd no.. hence sufficient.. II is not sufficient.. i hope it was of some help to those asking how n is odd..
_________________

Is Sn=n(n+1)/2 odd ?; Sn= sum of consecutive positive no.s

stat1: Sn/n is odd => n(n+1)/2n is odd or (n+1)/2 is odd or (n+1) is even and => n is odd or Sn = n(odd) * (n+1)/2 (odd) = odd suff.

stat2: n is odd => n+1 is even and n(n+1) is even or Sn = n(n+1)/2 is even... suff.

ANS:- i think u r going wrong on stat2... eg if 3 nos are 1,2,3.. n is 3 ie odd however its sum is 6 which is even.. or 3 nos are 2,3,4.. n is 3 ie odd however its sum is 9 which is odd.. so not sufficient Sn=n(n+1)/2 is the sum of first consecutie positive nos .....here it is not given that they are first consecutie positive nos but only that they are consecutie positive nos, where Sn = n(2a+(n-1)d)/2
_________________

Sum = k + (k+1) + (k +2) +.... + (k+n-1) = \(\frac{n(k+k+n-1)}{2}\)

=> Sum = \(\frac{n(2k+n-1)}{2}\)

1.) Sum/n = odd

=> Sum = n*odd..insufficient..(depends on n.)

2.) n is odd.. Sum = \(\frac{n(2k+n-1)}{2}\) \(=> Sum = odd * \frac{(even)}{2}\) Now, Even/2 can be odd or even..we can not be sure..insufficient..

combining both..

n is odd..

=> Sum is odd ..hence, C

(1) S=n*odd, S can be odd or even - generally right. But here, we have consecutive positive integers and here if S is even average is always decimal, if S is odd average can be even or odd, so if average=S/n is not decimals already means that S is odd.

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

Show Tags

06 Sep 2009, 17:55

1

This post received KUDOS

Bunuel wrote:

gmate2010 wrote:

Let z1 = k; z2 = k + 1 .....zn = k + n - 1

Sum = k + (k+1) + (k +2) +.... + (k+n-1) = \(\frac{n(k+k+n-1)}{2}\)

=> Sum = \(\frac{n(2k+n-1)}{2}\)

1.) Sum/n = odd

=> Sum = n*odd..insufficient..(depends on n.)

2.) n is odd.. Sum = \(\frac{n(2k+n-1)}{2}\) \(=> Sum = odd * \frac{(even)}{2}\) Now, Even/2 can be odd or even..we can not be sure..insufficient..

combining both..

n is odd..

=> Sum is odd ..hence, C

(1) S=n*odd, S can be odd or even - generally right. But here, we have consecutive positive integers and here if S is even average is always decimal, if S is odd average can be even or odd, so if average=S/n is not decimals already means that S is odd.

So A

hmmm..i forgot to apply the property of sum of consecutive integers...Thanks for correcting me..

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

Show Tags

02 Mar 2010, 09:46

1

This post received KUDOS

swethar wrote:

Please solve (with explanation):

If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd.

st 1) [(z1+z2+z3+...zn)/n] is the avg arithmetic mean of the series - for consequetive numbers, if the total number is even, then mean is the avg of the middle two numbers(which is not an interger), or if the total number is odd, then mean is the middle number. As its given mean is an odd interger, the middle number is an odd integer and we will have the same number of positive integers to the right of mean as to the left of mean. and the sum of the remaining integers except mean will be even. ( as for every odd number to the right of mean, there would be an odd number to the left of mean). So the sum of all the numbers in the series is odd. Sufficient

st 2) n is odd - sum could be even if the middle number(mean/median) is even [3,4,5] or sum could be odd if the middle number(mean/median) is odd [6,7,8] Not sufficient

If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd.

There is an important property of \(n\) consecutive integers: • If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

(1) \(\frac{z_1+z_2+z_3+...z_n}{n}=odd\), as the result of division the sum over the number of terms n is an integer, then n must be odd --> \(z_1+z_2+z_3+...z_n=odd*n=odd*odd=odd\). Sufficient.

(2) \(n\) is odd. Sum can be odd as well as even. Not sufficient.

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

Show Tags

03 Oct 2014, 06:05

swethar wrote:

If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?

(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd.

A.

1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.

If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?

(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd.

A.

1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

Show Tags

03 Oct 2014, 09:31

Bunuel wrote:

thefibonacci wrote:

swethar wrote:

If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?

(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd.

A.

1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.

1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.

Thanks Bunuel. But that would not make the sum odd, still. What am I missing here?

Sorry, but don't know what are you trying to prove there? What's your question? Do you get that n is even? Or ...? If you make last term a + n - 1 instead of a + n you'll get that n is odd.
_________________

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

Show Tags

15 Sep 2017, 04:02

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.
_________________