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

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..

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..

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

A

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.

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.

2) n = odd
1+2+3 = 6 (even)
1+2+3+4+5 = 15 (odd)
insufficient.

The last term would be a + n - 1, not a + n.

Given, z1, z2, z3, ..., znz1, z2, z3, ..., zn is a sequence of consecutive positive integers.

From (1) we have:
(z1+z2+z3+…+zn)/n = odd
This is only possible when (z1+z2+z3+…+zn) = odd.
Because we know, odd / odd = odd, even / odd = even and odd / even = Not an integer
So, statement (1) is sufficient.

From (2) we have n is odd
Say n = 3 then we have following possibilities
a. Odd + even + odd (as they are consecutive) = even
b. Even + odd + even = odd
So, st. 2 is not sufficient.

Hence answer choice A.

Statement 1 Alone:

This tells us the mean is an odd number. In arithmetic sequences, the mean is equal to the median, so this tells us the median is an odd number as well. If the median is an integer for consecutive numbers, then we must have an odd number of terms. If we had an even number of terms, the median would be X.5 (e.g. {2, 3} has a mean/median of 2.5).

Thus the sum is Odd*Odd = Odd. Sufficient.

Statement 2 Alone:

We don't know if the mean/median is even or odd. Then the sum could be even or odd. Insufficient.

Answer: A

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