Solution:

Since $k(k-1)$ is even, $n_k = \frac{k(k-1)}{2}$ is an integer and $\cos\frac{k(k-1)\pi}{2} = \cos(n_k \pi) = (-1)^{n_k}.$ The parity of the triangular number $n_k$ depends only on $k \bmod 4:$ it is even for $k \equiv 0, 1 \pmod 4$ and odd for $k \equiv 2, 3 \pmod 4.$ So $A_k = n_k$ when $k \equiv 0, 1 \pmod 4$ and $A_k = -n_k$ when $k \equiv 2, 3 \pmod 4.$

Group the $80$ terms into $20$ consecutive blocks of four starting at $k = 19 \equiv 3 \pmod 4.$ Using $n_{j+1} - n_j = j,$ each block with $k \equiv 3 \pmod 4$ collapses: $$A_k + A_{k+1} + A_{k+2} + A_{k+3} = (n_{k+1} - n_k) - (n_{k+3} - n_{k+2}) = k - (k + 2) = -2.$$

The total is $20 \cdot (-2) = -40,$ so the requested absolute value is $40.$