MHB Finding Mean of a Set of Abstract Numbers

AI Thread Summary
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.
cmkluza
Messages
118
Reaction score
1
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!
 
Mathematics news on Phys.org
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!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top