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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..

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

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13 Jul 2016, 05:46

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