The code looks like this:
== code(c) |

i = l
while i % k != 0 and i < r:
  sum += a[i]
  lo += 1
while i + k < r:
   lo += k
   sum += s[lo/k]
while lo + 1 <= r:
   lo += 1
   sum += a[lo]

i = lo
while (i + 1) % k != 0 and i <= hi:
  sum += A[i]
  i += 1
while i + k <= hi:
   sum += S[i/k]
   i += k
while i <= hi:
   sum += A[i]
   i += 1

==
The query takes less than $k + n/k + k = 2k + n/k$ time. $2k + n/k$ is minimized when $k$ is $O(sqrt(n))$. For $k = sqrt(n)$ the query takes $O(sqrt(n))$ time.