rrobiinn wrote:...yours has made it complex for me to understand. Could you make your explanation bit more easy to comprehend?
Let me try.
Statement 1: Average of any even number of consecutive integers will always be the average of the central two numbers, i.e. a fraction.
For example, say we have four consecutive integers: a, (a + 1), (a + 2), and (a + 3)
Average of this four consecutive integers = (4a + 6)/4 = (2a + 3)/2
Which is same as the average of central two numbers = [(a + 1) + (a + 2)]/2 = (2a + 3)/2 = (a + 1.5) --> A fraction
Let's take another example
Say we have ten consecutive integers: a, (a + 1), (a + 2), (a + 3), (a + 4), (a + 5), (a + 6), (a + 7), (a + 8), and (a + 9)
Average of this four consecutive integers = (10a + 45)/10 = (2a + 9)/2
Which is same as the average of central two numbers = [(a + 4) + (a + 5)]/2 = (2a + 9)/2 = (a + 4.5) --> A fraction
The average will always be a fraction as the sum of central two integers will always be odd as they are consecutive integers.
Hence, the average cannot be equal to 1.
Sufficient
Statement 2: Sum of n integers = s and 0 < s < n ---> 0 < s/n < 1
We know, s is less than n. Hence, the average, i.e. the ratio of s and n cannot be equal to n. This is just saying in words. I have shown it mathematically as 0 < s < n implies 0 < s/n < 1
Hence, average of those n consecutive integers = s/n < 1
Sufficient
I could've picked some number and shown that the statements do not help to conclude that average is not equal to 1 but that does not mean that they won't for any other set of numbers. If the question was "
Is the average of n consecutive integers always equal to 1?" Then by picking some numbers and showing that average is not equal to 1, we could've concluded for sure that the average is not
always equal to 1.
Picking number is great for showing exceptions or different scenarios but we can't pick number and prove some general case.
