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Is the average of n consecutive integers equal to 1 ?
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Updated on: 05 Nov 2013, 02:19
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Is the average of n consecutive integers equal to 1 ? (1) n is even (2) if S is the sum of the n consecutive integers, then 0 < S < n Correct answer D
Hello Fluke and Everyone in this forum  can you please explain the conceptual way of to solve this and improve my understanding ?
My approach to solve this:
Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.
Statement 2: I understand what the statement means, but I am struggling to understand the concept.
My understanding is
Average = Sum/Number of terms Therefore, Sum = Average * Number of Terms Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.
So the answer is D. But I don't understand the Statement 2.
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Originally posted by enigma123 on 02 Jul 2011, 13:30.
Last edited by Bunuel on 05 Nov 2013, 02:19, edited 2 times in total.
Edited the question.



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Re: Average of Consecutive Integers
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02 Jul 2011, 13:52
statement 1 is fairly clear as you said statement 2 says : S is positive and S>n.....(eq) now as you said sum = avg*n = S subtitute value of S in equation we get => avg*n>n now since n cant be negative, we can cancel n out on both sides without a change of sign which gives me =>avg>1 so it succesfully answers NO! to is avg=1? hope its clear...
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Re: Average of Consecutive Integers
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02 Jul 2011, 13:55
enigma123 wrote: Is the average of \(n\) consecutive integers equal to \(1\)?
1) \(n\) is even 2) If \(S\) is the sum of the \(n\) consecutive integers then \(0<S>n\).
2) To get an average=1; the Sum of elements must be equal to number of elements. \(If S=n; \frac{S}{n}=1\); If S is NOT equal to n, the average can not be 1. Sum=100; n=100; S/n=100/100=1 Sum=1; n=1; S/n=1/1=1 Sum=10; n=9; S/n=10/9=1.##(NOT EQUAL TO 1) Sufficient.
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Re: Average of Consecutive Integers
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03 Jul 2011, 00:02
My apologies. The second statement should have read if S is the sum of the n consecutive integers, then 0<S<n.
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Re: Average of Consecutive Integers
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03 Jul 2011, 00:44
So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please?
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Re: Average of Consecutive Integers
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03 Jul 2011, 01:27
enigma123 wrote: So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please? s<n or s>n means \(s \ne n\). If s NOT EQUAL n, the average can't be 1.
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Re: Average of Consecutive Integers
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03 Jul 2011, 01:29
Many thanks Fluke. Appreciate all your help.
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Re: Is the average of n consecutive integers equal to 1 ? 1) n
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05 Nov 2013, 02:11
Dear enigma123, Please edit stm2 as you posted it will be "if S is the sum of the n consecutive integers, then 0<S<n".
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Re: Is the average of n consecutive integers equal to 1 ? 1) n
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05 Nov 2013, 02:22



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Re: Is the average of n consecutive integers equal to 1 ?
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31 Jan 2014, 13:24
enigma123 wrote: Is the average of n consecutive integers equal to 1 ? (1) n is even (2) if S is the sum of the n consecutive integers, then 0 < S < n Correct answer D
Hello Fluke and Everyone in this forum  can you please explain the conceptual way of to solve this and improve my understanding ?
My approach to solve this:
Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.
Statement 2: I understand what the statement means, but I am struggling to understand the concept.
My understanding is
Average = Sum/Number of terms Therefore, Sum = Average * Number of Terms Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.
So the answer is D. But I don't understand the Statement 2. Is the average of n consecutive integers = 1? Statement 1 n is even If n is even then the average of consecutive integers will not be an integer at all so it won't be 1 for sure Just to illustrate n = 2, the avarege of 1 and 2 is 1.5 n=4, the average of 1,2,3 and 4 is 2.5 etc... Sufficient Statement 2 We can manipulate this to 0<s/n<1 So we are told that the average is in fact less than 1 Therefore sufficient as well D



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Re: Is the average of n consecutive integers equal to 1 ?
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01 Feb 2014, 03:03
+1 D Stmt1: n is even. let n=4, and 4 consecutive integers can be a1,a,a+1,a+2. sum=a1+a+a+1+a+2 = 4a+2 their average is (4a+2)/4 a+1/2 since a is integer and adding a fraction to a. mean can never be 1. SUFF Stmt2: 0<S<n. for average to be 1 S must be equal to n. Hence Stmt2 is also SUFF.
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Re: Is the average of n consecutive integers equal to 1 ?
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26 Apr 2016, 18:24
stmt 1 says that n is even
we can try with 2 integers: 2 and 3 , the average is 2.5 we can try with 4 integers: 2, 3, 4, 5, the average is 3.5
there is a rule that the average of a consecutive even set of numbers will not result in an integer.
Sufficient.
stmt2
0 < S < n, where S = Sum of terms and n = number of terms.
S = n * average of set
manipulate this formula to Average = S/n
If S < n, then S/n cannot be equal to 1.
Sufficient.



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Re: Is the average of n consecutive integers equal to 1 ?
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26 Apr 2016, 21:45
if average has to be = 1 , sum of elements must be equal to n . E G. 0,1,2 ; 1,0,1,2,3 .
if number of elements is odd in consecutive numbers, middle element= average. if number of elements is even , average = sum of middle two elements/2 > odd number/2 so average would not be an integer.
statement 1: n is even therefore n can not be equal to 1 so it gives a definite Ans statement 2: 0<S<n but for average to be 1, S must be equal to n so it gives a definite Ans
therefore D.



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Is the average of n consecutive integers equal to 1 ?
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20 Dec 2016, 16:55
Great Question. Here is what i did in this one >
We are asked if the mean of n consecutive is an integer or not.
N consecutive integer => Evenly spaced set =>Mean = Median =Average of the first and the last term
Mean for N consecutive integers can take only 2 forms => x => For N = odd x.5=> for N=even
for some integer x
Statement 1> N is even
So mean will never be an integer(Because median will never be an integer) Hence mean can never be "one". Hence Sufficient
Statement 2> Sum=> S And 0<S<N
\(Mean = \frac{Sum}{#}\)
Mean = \(\frac{S}{N}\)
As \(0<S<N\)=>
\(\frac{S}{N}\) will be a decimal between (0,1)
Hence mean will never be one. Hence Sufficient
Hence D
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Re: Is the average of n consecutive integers equal to 1 ?
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15 Feb 2017, 11:43
fivedaysleft wrote: statement 1 is fairly clear as you said
statement 2 says : S is positive and S>n.....(eq) now as you said sum = avg*n = S
subtitute value of S in equation we get => avg*n>n now since n cant be negative, we can cancel n out on both sides without a change of sign which gives me =>avg>1 so it succesfully answers NO! to is avg=1?
hope its clear... If Set = { 1,0,1,2} then Sum = 2 and n = 4, so S < n. Isn't this a valid scenario. In this case S < n S2 is valid.
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Re: Is the average of n consecutive integers equal to 1 ?
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15 Feb 2017, 12:36
coolkl wrote: fivedaysleft wrote: statement 1 is fairly clear as you said
statement 2 says : S is positive and S>n.....(eq) now as you said sum = avg*n = S
subtitute value of S in equation we get => avg*n>n now since n cant be negative, we can cancel n out on both sides without a change of sign which gives me =>avg>1 so it succesfully answers NO! to is avg=1?
hope its clear... If Set = { 1,0,1,2} then Sum = 2 and n = 4, so S < n. Isn't this a valid scenario. In this case S < n S2 is valid. Yes, that's a valid scenario. But is the average = 1?
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Re: Is the average of n consecutive integers equal to 1 ?
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06 Mar 2017, 06:47
Hi Bunuel , Could you post the detail solution for this question . Regards,
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Re: Is the average of n consecutive integers equal to 1 ?
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06 Mar 2017, 09:05
abhisheknandy08 wrote: Hi Bunuel ,
Could you post the detail solution for this question .
Regards, Is the average of n consecutive integers equal to 1 ?Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets: {1} {0, 1, 2} {1, 0, 1, 2, 3} So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd. (1) n is even > n is not odd > 1 cannot be the average. Sufficient. (2) if S is the sum of the n consecutive integers, then 0 < S < n > divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient. Answer: D. Hope it's clear.
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