1. **Stating the problem:** A person deposits money in a bank every month starting with 13000 in the first month, 14500 in the second month, 16000 in the third month, and so on, increasing by a fixed amount each month. We need to find the total amount deposited at the end of two years (24 months). 2. **Identify the sequence type:** The deposits form an arithmetic sequence where the first term $a_1 = 13000$ and the second term $a_2 = 14500$. 3. **Find the common difference $d$:** $$d = a_2 - a_1 = 14500 - 13000 = 1500$$ 4. **Number of terms $n$:** Since the deposits are monthly for two years, $n = 24$. 5. **Formula for the sum of an arithmetic series:** $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 6. **Substitute the values:** $$S_{24} = \frac{24}{2} (2 \times 13000 + (24-1) \times 1500)$$ 7. **Calculate inside the parentheses:** $$2 \times 13000 = 26000$$ $$(24-1) = 23$$ $$23 \times 1500 = 34500$$ 8. **Sum inside the parentheses:** $$26000 + 34500 = 60500$$ 9. **Calculate the total sum:** $$S_{24} = 12 \times 60500 = 726000$$ **Final answer:** The total amount deposited at the end of two years is **726000**.