Why arithmetic progression formula $S_n = (a_1 + a_n)*n/2$ works with uneven number of integer members?

Let's consider arithmetic progression with integer numbers.

Arithmetic progression sum $S_n = (a_1 + a_n)*n/2$, where $a_n=a_1+d(n-1)$

So $S_n = (2*a_1 + d(n-1))*n/2 = a_1*n + d(n-1)*n/2$

I cannot understand, why it always happens that $d(n-1)*n/2$ is always an integer number? So that $S_n$ is also always an integer.

Besides, everything seems clear with even number of progression members: 1 + 2 + 3 + 4 + 5 + 6 = (1+6) + (2+5) + (3+4) = (1+6)* (6 members / 2)

But if number of progression members is not even (1 + 2 + 3), it is unclear why $S_n = (a_1 + a_n)*n/2$ formula works perfectly! Because (3 members / 2) is not an integer!

Answer

$\frac{d(n-1)n}{2}$ is always an integer because $\frac{(n-1)n}{2}$ is always an integer. This is simply because $n$ and $n-1$ differ by $1$, and so atleast one of them must be even.