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If z1, z2, z3, ..., zn is a series of consecutive positive

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If z1, z2, z3, ..., zn is a series of consecutive positive [#permalink] New post 19 Aug 2009, 06:35
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If z1, z2, z3, ..., zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

(1) (z1+z2+z3+...+zn)/ n is an odd integer
(2) n is odd
[Reveal] Spoiler: OA
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Re: Series [#permalink] New post 19 Aug 2009, 07:06
TheRob wrote:
Hi , I have always had trouble with this thype of questions, would you please explain me how to solve it and how to get better at it?

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

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

2) n is odd


for (2), suppose n=3,
if z={1,2,3}, sum(z)=6
if z={2,3,4}, sum(z)=9
therefore 2) is nsf.

for (1),
(z1+z2+z3+...+zn)/ n =m, and m is an odd integer, therefore n is odd
consequently, sum(z)=n*m is an odd figure.

Answer is A.
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Re: Series [#permalink] New post 19 Aug 2009, 11:03
How do you find out that n is odd?
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Re: Series [#permalink] New post 06 Sep 2009, 09:49
TheRob wrote:
Hi , I have always had trouble with this thype of questions, would you please explain me how to solve it and how to get better at it?

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

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

2) n is odd


for the sum to be odd, n = odd , odd numbers included are odd in number

from 1

sum is even or odd., n is odd OR EVEN..........insuff

from 2

insuff

BOTH

n is odd still sum is surely odd ........C

Last edited by yezz on 06 Sep 2009, 09:52, edited 1 time in total.
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Re: Series [#permalink] New post 06 Sep 2009, 10:23
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..
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Re: Series [#permalink] New post 06 Sep 2009, 11:10
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.

IMO D
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Re: Series [#permalink] New post 06 Sep 2009, 11:23
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
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Re: Series [#permalink] New post 06 Sep 2009, 16:23
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
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Re: Series [#permalink] New post 06 Sep 2009, 16:49
Expert's post
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
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Re: Series [#permalink] New post 06 Sep 2009, 16:55
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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..
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Re: Series [#permalink] New post 07 Sep 2009, 07:29
Agree, I made a mistake; it should be A
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Re: If z1, z2, z3, ..., zn is a series of consecutive positive [#permalink] New post 26 May 2014, 23:21
TheRob wrote:
If z1, z2, z3, ..., zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

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


As per the information

1. Average of number from Z1 to Zn is an odd integer.

In other words, (First Term+Last Term)/2= odd integer
FT+LT= Even integer

It is only possible if FT and LT are both odd or both even at the same time.

Now between x consecutive positive odd integers the number of terms is odd
between x consecutive positive even integers the number of terms is odd

so the sum of the terms= [(FT+LT)/2] X no. of terms from FT to LT inclusive
= odd integer X odd integer
= odd integer
1. Sufficient

2. Clearly not sufficient.

Ans. A.
Re: If z1, z2, z3, ..., zn is a series of consecutive positive   [#permalink] 26 May 2014, 23:21
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