Slope of a triangle where the sum of the height at each integer equals 1.

Question

I have a problem where I'm trying to determine a proof to get the slope of a triangle $m$ in terms of $n$ where $\sum_{i=1}^n h_i = 1$.

I think I stumbled apon the answer being $m = \frac{2}{n(n+1)}$ but I don't fully understand how I got there. Can anyone help explain? Thanks.

EDIT: Solved! For any possible future readers: $$m = h_1, \;\;h_i = ih_1$$ $$\sum_{i=1}^n h_i = \sum_{i=1}^n ih_1 = h_1\sum_{i=1}^n i = 1$$ $$h_1 (1+2+...+n) = 1$$ $$h_1\frac{n(n+1)}{2}=1$$ $$h_1 = \frac{2}{n(n+1)} $$

Answer 1

Hint: The slope is equal to $h_1$, by definition of slope. Express an arbitrary height $h_i$ in terms of $h_1$, then see what $\sum h_i=1$ says about $h_1$.