Mean Data Is Mean

$\text {Let's call the sum of the first n numbers } S_n \text { (and } S_0 = 0 \text { )}$

$\text { and the nth term of this arrangement } a_n .$

$\text {Since the mean of the first n numbers is n, therefore:}$

$\frac {S_n}{n} = n \iff S_n = n^2$

Now:

$a_n = S_n - S_{n - 1} = n^2 - (n - 1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1$

By substituting n= 1, 2, ..., 9, 10, we get our numbers:

1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.

Therefore, our answer should be:

$\boxed {\text {They are all odd numbers}}$

$\text{Let all the numbers be odd-numbers.} \\ \sum_{i=1}^{n} x_i \text{, where x is ascending odd number} \\ \text{Ascending odd number is in the form of AP, where} a=1, d=2 \text{So, }\\ x_i = a + (i-1)d = 1 + (i-1)2 = 1 + 2i - 2 = 2i - 1 \\ \implies \sum_{i=1}^{n} 2i - 1 \\ \implies \sum_{i=1}^{n} 2i - 1 = \sum_{i=1}^{n} 2i + \sum_{i=1}^{n} - 1 \\ = 2\sum_{i=1}^{n} i - n = 2 \dfrac{n(n+1)}{2} - n \\ = n(n+1) - n = n^2 + n - n = n^2 \\ \implies \sum_{i=1}^{n} 2i - 1 = n^2 \\ \text{Mean, }\bar{x} = \dfrac{\sum_{i=1}^{n} x_i}{n} = \dfrac{\sum_{i=1}^{n} 2i - 1}{n}\\ = \dfrac{n^2}{n} = n \\ \text{Therefore, mean of all n-consecutive odd numbers is no of odd numbers chosen.} \\ \text{But on taking }x_i \text{ as an even number, }x_i \text{ = 2n} \\ \text{Which would then give sum to n of x as } n(n+1) \\ \text{So the mean would then be } \dfrac{n(n+1)}{n} = n + 1 \neq n$

Therefore all must be odd numbers

No need to check anything. The mean of first number is 1 equal to number itself. So 1 is odd and only 1 option satisfies.

Let a, b, c,....j be the 10 numbers ordered. We have a/1 = 1, a+b/2 = 2, a+b+c/3 = 3, ..... The only way it works must be to have odd numbers because odd = odd, odd+odd = even, odd+odd+odd = odd, .... Alternatively it permits to have the divisibilities asked. For example, we couldn t have (even+even+even)/3 = 3 .