1. The problem states that the numbers 11, p, q, and 21/1/2 (which is 21.5) form an arithmetic progression (AP). 2. In an arithmetic progression, the difference between consecutive terms is constant. Let's denote this common difference as $d$. 3. The first term $a_1 = 11$. 4. The second term $a_2 = p = a_1 + d = 11 + d$. 5. The third term $a_3 = q = a_2 + d = (11 + d) + d = 11 + 2d$. 6. The fourth term $a_4 = 21.5 = q + d = (11 + 2d) + d = 11 + 3d$. 7. From step 6, we have $11 + 3d = 21.5$ 8. Solve for $d$: $$3d = 21.5 - 11 = 10.5$$ so $$d = \frac{10.5}{3} = 3.5$$ 9. Substitute $d$ back to find $p$: $$p = 11 + d = 11 + 3.5 = 14.5$$ 10. Substitute $d$ back to find $q$: $$q = 11 + 2d = 11 + 2 \times 3.5 = 11 + 7 = 18$$ **Final answer:** $$p = 14.5, \quad q = 18$$