1. Problem: The third, fifth, and seventeenth terms of an arithmetic progression (AP) are in geometric progression (GP). Find the common ratio of the GP. 2. Let the first term of the AP be $a$ and the common difference be $d$. 3. The $n$th term of an AP is given by the formula: $$a_n = a + (n-1)d$$ 4. The third term is $a_3 = a + 2d$, the fifth term is $a_5 = a + 4d$, and the seventeenth term is $a_{17} = a + 16d$. 5. Since these three terms are in GP, the ratio between consecutive terms is constant: $$\frac{a_5}{a_3} = \frac{a_{17}}{a_5}$$ 6. Substitute the terms: $$\frac{a + 4d}{a + 2d} = \frac{a + 16d}{a + 4d}$$ 7. Cross-multiply: $$(a + 4d)^2 = (a + 2d)(a + 16d)$$ 8. Expand both sides: $$a^2 + 8ad + 16d^2 = a^2 + 18ad + 32d^2$$ 9. Simplify by subtracting $a^2$ from both sides: $$8ad + 16d^2 = 18ad + 32d^2$$ 10. Rearrange terms: $$8ad + 16d^2 - 18ad - 32d^2 = 0$$ 11. Simplify: $$-10ad - 16d^2 = 0$$ 12. Factor out $-2d$: $$-2d(5a + 8d) = 0$$ 13. Since $d \neq 0$ (otherwise no progression), set: $$5a + 8d = 0 \implies a = -\frac{8}{5}d$$ 14. Find the common ratio $r$ of the GP: $$r = \frac{a_5}{a_3} = \frac{a + 4d}{a + 2d} = \frac{-\frac{8}{5}d + 4d}{-\frac{8}{5}d + 2d} = \frac{-\frac{8}{5} + 4}{-\frac{8}{5} + 2} = \frac{\frac{12}{5}}{\frac{2}{5}} = 6$$ 15. Final answer: The common ratio of the geometric progression is $6$. 16. Problem: A child receives pocket money doubling each day starting with 1 on day 1, 2 on day 2, 4 on day 3, etc., for 30 days. Find the total amount received. 17. This is a geometric progression with first term $a = 1$ and common ratio $r = 2$. 18. The sum of the first $n$ terms of a GP is: $$S_n = a \frac{r^n - 1}{r - 1}$$ 19. Substitute $a=1$, $r=2$, $n=30$: $$S_{30} = \frac{2^{30} - 1}{2 - 1} = 2^{30} - 1$$ 20. Calculate $2^{30} = 1073741824$, so: $$S_{30} = 1073741824 - 1 = 1073741823$$ 21. To the nearest million, the child receives $1073741823 \approx 1073742$ million. 22. Problem: Find the common ratio of a GP with first term 5 and sum to infinity 15. 23. Sum to infinity of a GP with $|r|<1$ is: $$S_\infty = \frac{a}{1 - r}$$ 24. Substitute $a=5$, $S_\infty=15$: $$15 = \frac{5}{1 - r} \implies 1 - r = \frac{5}{15} = \frac{1}{3}$$ 25. Solve for $r$: $$r = 1 - \frac{1}{3} = \frac{2}{3}$$ 26. Final answer: The common ratio is $\frac{2}{3}$. 27. Problem: The sum of the first two terms of a GP is 9, sum to infinity is 25, and common ratio is positive. Find $r$ and $a$. 28. Sum of first two terms: $$a + ar = a(1 + r) = 9$$ 29. Sum to infinity: $$S_\infty = \frac{a}{1 - r} = 25$$ 30. From sum to infinity: $$a = 25(1 - r)$$ 31. Substitute into sum of first two terms: $$25(1 - r)(1 + r) = 9$$ 32. Simplify: $$25(1 - r^2) = 9 \implies 1 - r^2 = \frac{9}{25} = 0.36$$ 33. Solve for $r^2$: $$r^2 = 1 - 0.36 = 0.64$$ 34. Since $r$ is positive: $$r = 0.8$$ 35. Find $a$: $$a = 25(1 - 0.8) = 25(0.2) = 5$$ 36. Final answer: $r = 0.8$, $a = 5$. 37. Problem: Bacteria doubles every 2 hours. Starting with 500 bacteria, find the number after 24 hours. 38. Number of doubling periods in 24 hours: $$\frac{24}{2} = 12$$ 39. Number of bacteria after 24 hours: $$N = 500 \times 2^{12} = 500 \times 4096 = 2048000$$ 40. Final answer: There will be 2048000 bacteria after 24 hours. 41. Problem: Hot tub temperature increases by 10% each hour. Starting at 75ºF, find temperature after 3 hours. 42. This is a geometric progression with $a=75$, $r=1.10$, $n=3$. 43. Temperature after 3 hours: $$T = 75 \times 1.10^3 = 75 \times 1.331 = 99.825$$ 44. Rounded to nearest tenth: $$99.8$$ 45. Final answer: The temperature after 3 hours is 99.8ºF. 46. Problem: Find the sum of interior angles of a dodecagon (12 sides). 47. The sum of interior angles of an $n$-sided polygon is: $$S = 180(n - 2)$$ 48. Substitute $n=12$: $$S = 180(12 - 2) = 180 \times 10 = 1800$$ 49. Final answer: The sum of interior angles of a dodecagon is 1800º.