1. **Problem 1: Height of the bottom drawer in a geometric progression** The height of the drawers forms a geometric progression with first term $a_1=6$ inches and common ratio $r=1.5$. There are 4 drawers. 2. **Formula:** The $n$th term of a geometric sequence is given by $$a_n = a_1 \times r^{n-1}$$ 3. **Calculate the height of the 4th drawer:** $$a_4 = 6 \times 1.5^{4-1} = 6 \times 1.5^3$$ 4. Calculate $1.5^3$: $$1.5^3 = 1.5 \times 1.5 \times 1.5 = 3.375$$ 5. Multiply: $$a_4 = 6 \times 3.375 = 20.25$$ 6. **Answer:** The bottom drawer is 20.25 inches high, which corresponds to option D. --- 7. **Problem 2: Chad's salary in year 5 with 3% annual raise** Initial salary $S_1 = 24000$, raise rate $r = 1 + 0.03 = 1.03$, find salary in year 5. 8. **Formula:** Salary in year $n$ is $$S_n = S_1 \times r^{n-1}$$ 9. Calculate $S_5$: $$S_5 = 24000 \times 1.03^{4}$$ 10. Calculate $1.03^4$: $$1.03^4 = 1.03 \times 1.03 \times 1.03 \times 1.03 = 1.1255$$ (approx) 11. Multiply: $$S_5 = 24000 \times 1.1255 = 27012$$ (approx) 12. **Answer:** Chad makes approximately 27012 in year 5, option C. --- 13. **Problem 3: Number of stores 18 years after 1987** Given a geometric sequence with data points: - Year 1: 60 stores - Year 8: 215 stores - Year 9: 258 stores - Year 14: 642 stores 14. **Find common ratio $r$:** Using years 1 and 8: $$a_8 = a_1 \times r^{7}$$ $$215 = 60 \times r^{7}$$ $$r^{7} = \frac{215}{60} = 3.5833$$ 15. Take 7th root: $$r = \sqrt[7]{3.5833} = 1.2$$ (approx) 16. **Find $a_{18}$:** $$a_{18} = 60 \times 1.2^{17}$$ 17. Calculate $1.2^{17}$: $$1.2^{17} \approx 22.18$$ 18. Multiply: $$a_{18} = 60 \times 22.18 = 1331$$ (approx) 19. **Answer:** The company owned approximately 1331 stores 18 years after 1987, option C.