1. **Problem Statement:** We are given that the sum of the first $n$ terms of two arithmetic progressions (APs) are in the ratio $3n + 31 : 5n - 3$. We need to find the ratio of their 9th terms. 2. **Formula for sum of first $n$ terms of an AP:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ where $a$ is the first term and $d$ is the common difference. 3. **Let the first AP have first term $a_1$ and common difference $d_1$, and the second AP have first term $a_2$ and common difference $d_2$.** 4. The sums of the first $n$ terms are: $$S_n^{(1)} = \frac{n}{2} [2a_1 + (n-1)d_1]$$ $$S_n^{(2)} = \frac{n}{2} [2a_2 + (n-1)d_2]$$ 5. Given the ratio: $$\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{3n + 31}{5n - 3}$$ 6. Substitute the sums: $$\frac{\frac{n}{2} [2a_1 + (n-1)d_1]}{\frac{n}{2} [2a_2 + (n-1)d_2]} = \frac{3n + 31}{5n - 3}$$ 7. Simplify by canceling $\frac{n}{2}$: $$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 31}{5n - 3}$$ 8. This equality holds for all $n$, so the numerators and denominators must be linear expressions in $n$ that match the ratio. Write: $$2a_1 + (n-1)d_1 = A n + B$$ $$2a_2 + (n-1)d_2 = C n + D$$ 9. Express $2a_1 + (n-1)d_1$ as: $$2a_1 + (n-1)d_1 = 2a_1 - d_1 + d_1 n = d_1 n + (2a_1 - d_1)$$ Similarly for the second AP: $$2a_2 + (n-1)d_2 = d_2 n + (2a_2 - d_2)$$ 10. So the ratio is: $$\frac{d_1 n + (2a_1 - d_1)}{d_2 n + (2a_2 - d_2)} = \frac{3n + 31}{5n - 3}$$ 11. Equate coefficients: $$\frac{d_1}{d_2} = \frac{3}{5}$$ $$\frac{2a_1 - d_1}{2a_2 - d_2} = \frac{31}{-3} = -\frac{31}{3}$$ 12. We want the ratio of the 9th terms: $$T_9^{(1)} = a_1 + 8 d_1$$ $$T_9^{(2)} = a_2 + 8 d_2$$ 13. Express $a_1$ and $a_2$ in terms of $d_1$ and $d_2$: From step 11 second equation: $$\frac{2a_1 - d_1}{2a_2 - d_2} = -\frac{31}{3}$$ Let $d_2 = d_2$, and $d_1 = \frac{3}{5} d_2$ from step 11 first equation. 14. Write: $$2a_1 - d_1 = -\frac{31}{3} (2a_2 - d_2)$$ 15. Solve for $a_1$: $$2a_1 = d_1 - \frac{31}{3} (2a_2 - d_2)$$ $$a_1 = \frac{d_1}{2} - \frac{31}{6} (2a_2 - d_2)$$ 16. Now find the ratio: $$\frac{T_9^{(1)}}{T_9^{(2)}} = \frac{a_1 + 8 d_1}{a_2 + 8 d_2}$$ Substitute $a_1$ and $d_1$: $$= \frac{\frac{d_1}{2} - \frac{31}{6} (2a_2 - d_2) + 8 d_1}{a_2 + 8 d_2}$$ $$= \frac{\frac{3}{5} \frac{d_2}{2} - \frac{31}{6} (2a_2 - d_2) + 8 \times \frac{3}{5} d_2}{a_2 + 8 d_2}$$ 17. Simplify numerator: $$\frac{3}{10} d_2 - \frac{31}{6} (2a_2 - d_2) + \frac{24}{5} d_2$$ 18. Combine $d_2$ terms: $$\frac{3}{10} d_2 + \frac{24}{5} d_2 = \frac{3}{10} d_2 + \frac{48}{10} d_2 = \frac{51}{10} d_2$$ 19. So numerator: $$\frac{51}{10} d_2 - \frac{31}{6} (2a_2 - d_2)$$ 20. Expand: $$\frac{51}{10} d_2 - \frac{31}{6} \times 2 a_2 + \frac{31}{6} d_2 = \frac{51}{10} d_2 + \frac{31}{6} d_2 - \frac{62}{6} a_2$$ 21. Combine $d_2$ terms: $$\frac{51}{10} d_2 + \frac{31}{6} d_2 = \frac{306}{60} d_2 + \frac{310}{60} d_2 = \frac{616}{60} d_2 = \frac{154}{15} d_2$$ 22. Numerator is: $$\frac{154}{15} d_2 - \frac{62}{6} a_2$$ 23. Denominator is: $$a_2 + 8 d_2$$ 24. To find the ratio, choose $a_2$ and $d_2$ to simplify. Let $d_2 = 15$ (to clear denominator 15) and $a_2 = 6$ (to clear denominator 6). 25. Substitute: Numerator: $$\frac{154}{15} \times 15 - \frac{62}{6} \times 6 = 154 - 62 = 92$$ Denominator: $$6 + 8 \times 15 = 6 + 120 = 126$$ 26. Ratio: $$\frac{92}{126} = \frac{46}{63}$$ 27. This ratio is approximately 0.73, which is not exactly any of the options. Check if options are simplified ratios. 28. Alternatively, check the options given: A. 1:1 = 1 B. 1:2 = 0.5 C. 2:1 = 2 D. 1:3 = 0.333... 29. Our ratio is about 0.73, closest to 1:1 but not exact. Re-examine step 11: the constant term ratio is negative, so the ratio of 9th terms might be negative or different. 30. Another approach: Since the ratio of sums is linear in $n$, the 9th term ratio is the ratio of coefficients of $n$ in sums. 31. The $n$th term $T_n = S_n - S_{n-1}$. 32. Calculate $T_n^{(1)}$: $$T_n^{(1)} = S_n^{(1)} - S_{n-1}^{(1)} = (3n + 31) - (3(n-1) + 31) = 3n + 31 - 3n + 3 - 31 = 3$$ 33. Calculate $T_n^{(2)}$: $$T_n^{(2)} = S_n^{(2)} - S_{n-1}^{(2)} = (5n - 3) - (5(n-1) - 3) = 5n - 3 - 5n + 5 + 3 = 5$$ 34. So the $n$th terms are constant: $T_n^{(1)} = 3$, $T_n^{(2)} = 5$ for all $n$. 35. Therefore, the ratio of 9th terms is: $$\frac{T_9^{(1)}}{T_9^{(2)}} = \frac{3}{5}$$ 36. None of the options exactly match $3:5$. But the closest is option B (1:2) or option A (1:1). Since $3:5$ is closer to $1:2$ than $1:1$, the answer is B. **Final answer:** The ratio of the 9th terms is $3:5$, which corresponds approximately to option B (1:2).