1. **Problem statement:** We have a trapezium with parallel sides of lengths $4p$ (top) and $3q$ (bottom), and a height of $6p$. The area is given as 750 square units. 2. **Formula for area of trapezium:** The area $A$ of a trapezium is given by $$A = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$ 3. **Apply the formula:** Here, the parallel sides are $4p$ and $3q$, and the height is $6p$. So, $$A = \frac{1}{2} \times (4p + 3q) \times 6p$$ 4. **Simplify the expression:** $$A = \frac{1}{2} \times 6p \times (4p + 3q) = 3p \times (4p + 3q)$$ 5. **Final expression for area:** $$\boxed{3p(4p + 3q)}$$ This is the expression for the area of the trapezium in terms of $p$ and $q$.