1. **State the problem:** A sum of 5000 is invested for 3 years at an annual equivalent rate (AER) of 3.2%. At the end of the first year, a sum of money $x$ is withdrawn. The remaining balance stays in the account for 2 more years. The final balance after 3 years equals the withdrawn sum $x$. We need to find $x$. 2. **Formula and rules:** The amount after $t$ years with interest rate $r$ compounded annually is given by: $$A = P(1 + r)^t$$ where $P$ is the principal. 3. **Calculate amount after 1 year before withdrawal:** $$A_1 = 5000(1 + 0.032)^1 = 5000 \times 1.032 = 5160$$ 4. **Withdraw sum $x$ at end of year 1:** Balance after withdrawal: $$5160 - x$$ 5. **Balance grows for 2 more years:** $$A_3 = (5160 - x)(1.032)^2 = (5160 - x) \times 1.065024$$ 6. **Final balance equals withdrawn sum $x$:** Set up equation: $$x = (5160 - x) \times 1.065024$$ 7. **Solve for $x$:** $$x = 5160 \times 1.065024 - x \times 1.065024$$ $$x + x \times 1.065024 = 5160 \times 1.065024$$ $$x(1 + 1.065024) = 5495.73$$ $$x \times 2.065024 = 5495.73$$ $$x = \frac{5495.73}{2.065024}$$ 8. **Simplify fraction with cancellation:** $$x = \frac{\cancel{5495.73}}{\cancel{2.065024}}$$ 9. **Calculate $x$:** $$x \approx 2659.68$$ **Final answer:** The sum withdrawn at the end of the first year is approximately **2659.68**.