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 350,000 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.

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..

It has been a fairly long time since I have posted here, but I definitely did not want to sign off without giving readers a quick update on my personal...