MHB Finding Mean of a Set of Abstract Numbers

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To find the mean of the set a, 2a, 3a, ..., na, the formula is derived as \frac{a(n+1)}{2}. The mean is calculated by summing the series and dividing by the number of terms, leading to the expression a(\frac{1+2+3+...+n}{n}). The sum of the first n integers, 1 to n, can be simplified using the formula (n(n+1))/2, which helps in deriving the mean. Understanding arithmetic progression is crucial for simplifying the sum and reaching the final expression. The discussion concludes with appreciation for the clarity gained in solving the problem.
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Hello, I'm having a little trouble figuring out the following problem:

Consider the set of number a, 2a, 3a, ..., na where a and n are positive integers.

(i) Show that the expression for the mean of this set is \frac{a(n+1)}{2}.

So far the only work I've been able to muster up is:

Mean = \frac{a+2a+3a+...+na}{n} = \frac{a(1+2+3+...+n)}{n} = a(\frac{1+2+3+...}{n}+1) = \frac{a+2a+3a+...}{n}+a

I'm not sure what to do with the indefinitely large sum of numbers that are involved with a in this problem, and I'm not really sure how to configure the problem into the simplified expression for mean shown in the problem.

Any help will be greatly appreciated, thanks!
 
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Read about arithmetic progression. In this case, it is sufficient to note that $1+\dots+n=(1+n)+(2+(n-1))+\dots$ where in the second sum the number of terms is $n/2$ and each term equals $n+1$. Consider what $n/2$ means more precisely in this context when $n$ is even and when it is odd.
 
Evgeny.Makarov said:
Read about arithmetic progression. In this case, it is sufficient to note that $1+\dots+n=(1+n)+(2+(n-1))+\dots$ where in the second sum the number of terms is $n/2$ and each term equals $n+1$. Consider what $n/2$ means more precisely in this context when $n$ is even and when it is odd.

Thanks a bunch for your help. Sorry I wasn't able to reply in a while, but it makes much more sense now. Thanks again!
 
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