Get ascending sum on 100% based on a number For each number I want to get the array of number ascending cumulative
For example :
$$100 \% = 45 \% + 55 \%$$
$$100 \% = 23.\overline{33} \% + 33.\overline{33} \% + 43.\overline{33} \%$$
$$100 \% = 10\% + 20\% + 30\% + 40\% $$
$$100 \% = a \%+b\%+c\%+d\%+e\%+f\%+g\%+h\% \ ??$$
What is the formula?
Thanks
 A: For a given $n$ you want the sum
$$a+(a+d)+\dots+(a+(n-1)d)=100$$
$$\frac{n}2(2a+(n-1)d)=100$$
$$a=\frac{100}n-\frac{(n-1)d}2$$
Hence we can choose any common difference $d$ and $a$ is given above. 
If you also want the difference between $0$ and the first term to be the same as the difference between the second and first terms, we need $a=d$. This gives us only one possible choice of $d$ namely
$$a=d=\frac{200}{n(n+1)}$$
But the difference between $100$ and the last term will not necessarily be the same. For example, choosing $n=3$ we get
$$a=d=\frac{50}{3}$$
Hence the sum becomes
$$100=\frac{50}3+\frac{100}3+\frac{150}3$$
A: Yo seem to want to get $n$ numbers in arithmetic progression such that their sum is $100$, so mean $\frac{100}{n}$, and each of the $n-1$ steps is $10$
So the lowest value is $\frac{100}{n} - 10\frac{n-1}{2} = \frac{100}{n} -5n+5$
and the highest value is $\frac{100}{n}  +10\frac{n-1}{2} = \frac{100}{n} +5n-5$
and with $n=8$ you get $$100 = -22.5 -12.5 -2.5 + 7.5+17.5+27.5+37.5+47.5$$
If you are not so fixed about the steps being $10$ and they are instead $d$ then the lowest value is $\frac{100}{n} - d\frac{n-1}{2}$ and the highest $\frac{100}{n} + d\frac{n-1}{2}$.  For example with $n=8$ and $d=3$ you could have $100 = 2+5+8+11+14+17+20+23$.
