1. **Problem Statement:** Calculate the amount to be deposited at the beginning of the 3rd year to buy a car costing 200000 after 3 years, given deposits of 80000 at the beginning of the 1st year and 40000 at the beginning of the 2nd year, with compound interest at 10% per annum. 2. **Formula Used:** The future value (FV) of a deposit compounded annually is given by: $$FV = P(1 + r)^n$$ where $P$ is the principal, $r$ is the interest rate, and $n$ is the number of years. 3. **Step 1: Calculate the future value of the first deposit (80000) after 3 years:** $$FV_1 = 80000(1 + 0.10)^3 = 80000(1.331) = 106480$$ 4. **Step 2: Calculate the future value of the second deposit (40000) after 2 years:** $$FV_2 = 40000(1 + 0.10)^2 = 40000(1.21) = 48400$$ 5. **Step 3: Let the amount to be deposited at the beginning of the 3rd year be $x$. It will earn interest for 1 year:** $$FV_3 = x(1 + 0.10) = 1.10x$$ 6. **Step 4: The sum of all future values must equal the cost of the car:** $$FV_1 + FV_2 + FV_3 = 200000$$ Substitute values: $$106480 + 48400 + 1.10x = 200000$$ 7. **Step 5: Solve for $x$:** $$1.10x = 200000 - 106480 - 48400 = 4520$$ $$x = \frac{4520}{1.10}$$ $$x = \frac{\cancel{4520}}{\cancel{1.10}} = 4109.09$$ 8. **Final Answer:** The amount to be deposited at the beginning of the 3rd year is approximately **4109.09**.