Proofreading Days C70141

1. **State the problem:** Liz reads 30 pages on the first day and increases the number of pages by 5 each subsequent day. We want to find how many days it will take her to read a total of 500 pages. 2. **Identify the sequence:** The number of pages read each day forms an arithmetic sequence with first term $a_1 = 30$ and common difference $d = 5$. 3. **Formula for the sum of an arithmetic sequence:** $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ where $S_n$ is the total pages read after $n$ days. 4. **Set up the equation:** We want $S_n \geq 500$. $$\frac{n}{2} (2 \times 30 + (n-1) \times 5) \geq 500$$ 5. **Simplify inside the parentheses:** $$2 \times 30 = 60$$ So, $$\frac{n}{2} (60 + 5n - 5) \geq 500$$ $$\frac{n}{2} (55 + 5n) \geq 500$$ 6. **Multiply both sides by 2 to clear denominator:** $$n (55 + 5n) \geq 1000$$ 7. **Distribute $n$:** $$55n + 5n^2 \geq 1000$$ 8. **Rewrite as a quadratic inequality:** $$5n^2 + 55n - 1000 \geq 0$$ 9. **Divide entire inequality by 5:** $$\cancel{5}n^2 + \cancel{5}11n - \cancel{5}200 \geq 0$$ $$n^2 + 11n - 200 \geq 0$$ 10. **Solve the quadratic equation $n^2 + 11n - 200 = 0$ using the quadratic formula:** $$n = \frac{-11 \pm \sqrt{11^2 - 4 \times 1 \times (-200)}}{2 \times 1}$$ $$n = \frac{-11 \pm \sqrt{121 + 800}}{2}$$ $$n = \frac{-11 \pm \sqrt{921}}{2}$$ 11. **Calculate approximate roots:** $$\sqrt{921} \approx 30.34$$ So, $$n = \frac{-11 + 30.34}{2} \approx \frac{19.34}{2} = 9.67$$ $$n = \frac{-11 - 30.34}{2} \approx \frac{-41.34}{2} = -20.67$$ 12. **Interpretation:** Since $n$ must be positive, we take $n \approx 9.67$ days. 13. **Since $n$ must be a whole number of days and she cannot finish in a fraction of a day, round up:** $$n = 10$$ days. **Final answer:** It will take Liz 10 days to complete proofreading 500 pages.