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I got a little bit confused...

Isn't it that for consecutive intergers the mean of the first+last are equal to the mean of the set of consecutive integers?

Even if they are consecutive odds, evens, or consecutive numbers that include negatives (like -2, -1, 0), the mean is always equal to the middle term.

So, both [1] and [2] actually say the same thing, that 2b=a+c. So, why cannot we assume that yes they are consecutive integers, and go for D?
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pacifist85
I got a little bit confused...

Isn't it that for consecutive intergers the mean of the first+last are equal to the mean of the set of consecutive integers?

Even if they are consecutive odds, evens, or consecutive numbers that include negatives (like -2, -1, 0), the mean is always equal to the middle term.

So, both [1] and [2] actually say the same thing, that 2b=a+c. So, why cannot we assume that yes they are consecutive integers, and go for D?

hi pacifist,

consecutive evens =2,4,6 or 6,8,10
consecutive odds= 1,3,5 or 5,7,9
and consecutive numbers = 2,3,4 or 5,6,7..

you are right the two statements mean the same that the centre term is avg of the 1st and 3rd number..
this basically means the 1st and 3rd numbers are at the same distance from middle no..
so options satisfying this are 1,2,3 or 2,4,6 or 2,6,10 and so on..
but as you can see it is not necessary they are consecutive numbers..
hope it cleared ur doubt.. :wink:
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What I mean is that, using alegbra, you get to 2B = A + C and to B = \(\frac{A + C}{2}\).

Do we need to test values because we only have variables? In other words, the "wise" way to approch this one would be to on-purpose-randomly choose 3 numbers that are non consecutive integers and the average of the first and last terms of which yields the middle?

E.g., 2, 7, 12 --> 12+2=14 and 14/2=7, but they are not consecutve.

I get the point of doing so, but it kind of sucks!
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I think the correct answer is D, if not ,
please inform me why?
thanx
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I think the correct answer is D, if not ,
please inform me why?
thanx

The answer should be E.

Statement 1 mentions that : 2b = a+c

Now, as there is no restriction on a,b,c apart from they being integers, we can assume them to be equal as well. Thus, a=b=c=0 satisfies this statement. But a=1,b=2 and c=3 also satisfies this statement. Thus, based on the first set of values, a,b,c are NOT consecutive integers while based on second set of values, a/b/c are consecutive integers. Thus this statement is not sufficient.

Statement 2 is a fancy way of again stating what statement 1 said.

Average of a,b,c = b

This means that b=(a+c)/2 , which is same as statement 1 and is thus insufficient.

Combining both 1 and 2, as both the statements give the same information, the given statements are still NOT sufficient and thus the answer is E.
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b=(a+c)/2 –> average=median –/–> consecutive integers
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Are the integers a, b and c consecutive?

(1) b - a = c - b
(2) The average of a, b and c is equal to b.

Concept: when integer is mentioned, immediately think in terms ZERO/POSITIVE/NEGATIVE/ODD/EVEN.
In this case Zero applies to statement One
Consecutive means always means integers that follow each other in order with common difference of ONE.

(1) b - a = c - b
Not sufficient all integers can be zero(as zero is also an integer) or they could be consecutive. Not Sufficient.

(2) The average of a, b and c is equal to b.
Not sufficient could be consecutive integers (common difference 1) or could be consecutive odd/even integers(common difference 2)

Combining (1) & (2) also all integers could be ZERO or they could be consecutive with a common difference ONE.
E
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