"Consider the function $f(x)=x$ on the interval $[0,a]$, where $a>0$. Let $P$ be an arbitrary partition of the interval $[0,a]$. Determine the Riemann sum associated with the midpoints $c_i$ of this partition."
Assuming that we are partitioning this area into sub-intervals $\Delta x$ of equal length, then $$\Delta x = \frac{a-0}{n}=\frac{a}{n}$$
and the midpoint of each subinterval is $c_i=\frac{a}{2n}$. The Riemann sum is then $$S(P,f,c_i)=\frac{a}{n}\cdot \Bigg[f\Bigg(\frac{a}{2n}\Bigg)+f\Bigg(\frac{3a}{2n}\Bigg)+\cdots+f\Bigg(a-\frac{a}{2n}\Bigg)\Bigg].(1)$$
Is this $$\frac{a}{n}\sum_{k=1}^nf\Bigg[\Big(2k-1\Big)\cdot\Big(\frac{a}{2n}\Big)\Bigg]$$
the proper generalized form of $(1)$?
Yes, it is. You could also maybe replace $f(x)$:
$$\frac an \sum_{k=1}^n (2k-1)\cdot\left(\frac{a}{2n}\right) \\=\frac{a^2}{n^2} \sum_{k=1}^n k -\frac{a^2}{2n^2}\sum_{k=1}^n1\\=\frac{a^2}{n}\cdot\frac{(n+1)}{2}-\frac{a^2}{2n}\\=\frac{a^2}{n} \left(\frac{n+1}{2}-\frac 12\right) \\=\frac{a^2}{2}$$