Project Euler Problem 1 Common Lisp

Project Euler Problem 1 

The idea here is to solve Problem 1 without using a modulus operator which I take to be more expensive than using addition and if.  So it adds every multiple of 5 except every third multiple, e.g., 5 + 10 + skip + 20 + 25 + skip.

(defun problem1 ()
  (+
    (loop for i from 3 below 1000 by 3 sum i)
    (loop for j from 5 below 1000 by 5
          and k = 1 then (if (= k 3) 1 (1+ k))
          unless (= k 3) sum j)))

There's a more efficient way to solve this without using a loop, but I didn't know about it until after I solved this, so at this point, for me, I consider it cheating.

Another solution avoiding a modulus operator but probably not any more efficient:

(defun problem1 ()
  (reduce #'+
          (union (loop for i below 1000 by 3 collect i)
                 (loop for i below 1000 by 5 collect i))))