1. Problem 32(a): An arithmetic progression (AP) has $n$ terms, the $n^{th}$ term is 4, and the common difference $d=2$. The sum of $n$ terms $S_n = -14$. Find $n$ and the sum of the first 20 terms. 2. Formula for the $n^{th}$ term of an AP: $$a_n = a + (n-1)d$$ where $a$ is the first term. 3. Formula for the sum of $n$ terms of an AP: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ 4. Using $a_n = 4$, substitute into the $n^{th}$ term formula: $$4 = a + (n-1)2$$ 5. Using $S_n = -14$, substitute into the sum formula: $$-14 = \frac{n}{2}[2a + (n-1)2]$$ 6. From step 4, express $a$: $$a = 4 - 2(n-1) = 4 - 2n + 2 = 6 - 2n$$ 7. Substitute $a$ into sum formula: $$-14 = \frac{n}{2}[2(6 - 2n) + 2(n-1)] = \frac{n}{2}[12 - 4n + 2n - 2] = \frac{n}{2}[10 - 2n]$$ 8. Simplify: $$-14 = \frac{n}{2}(10 - 2n) = 5n - n^2$$ 9. Rearrange: $$n^2 - 5n - 14 = 0$$ 10. Solve quadratic equation: $$n = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2}$$ 11. Possible values: $$n = \frac{5 + 9}{2} = 7 \quad \text{or} \quad n = \frac{5 - 9}{2} = -2$$ 12. Since $n$ must be positive, $n=7$. 13. Find $a$: $$a = 6 - 2(7) = 6 - 14 = -8$$ 14. Find sum of first 20 terms: $$S_{20} = \frac{20}{2}[2(-8) + (20-1)2] = 10[-16 + 38] = 10 \times 22 = 220$$ --- 15. Problem 32(b): Sum of first 6 terms $S_6=42$, ratio of 10th to 30th term is $1:3$. Find first term $a$ and 13th term $a_{13}$. 16. Use sum formula: $$42 = \frac{6}{2}[2a + (6-1)d] = 3[2a + 5d] = 6a + 15d$$ 17. So: $$6a + 15d = 42 \implies 2a + 5d = 14$$ 18. Use term ratio: $$\frac{a_{10}}{a_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3}$$ 19. Cross multiply: $$3(a + 9d) = a + 29d \implies 3a + 27d = a + 29d$$ 20. Rearrange: $$3a - a = 29d - 27d \implies 2a = 2d \implies a = d$$ 21. Substitute $a=d$ into $2a + 5d = 14$: $$2d + 5d = 14 \implies 7d = 14 \implies d = 2$$ 22. Then $a = 2$. 23. Find 13th term: $$a_{13} = a + 12d = 2 + 12 \times 2 = 2 + 24 = 26$$ --- 24. Problem 33: A vessel is an inverted cone with height $h=8$ cm and radius $r=5$ cm, filled with water. Lead shots are spheres of radius $0.5$ cm. When shots are dropped, one-fourth of water flows out. Find number of shots. 25. Volume of cone: $$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5)^2 (8) = \frac{1}{3} \pi 25 \times 8 = \frac{200}{3} \pi$$ 26. Volume of water displaced (one-fourth): $$\frac{1}{4} \times \frac{200}{3} \pi = \frac{50}{3} \pi$$ 27. Volume of one lead shot (sphere): $$V_s = \frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \pi \times 0.125 = \frac{1}{6} \pi$$ 28. Number of shots $n$: $$n = \frac{\text{displaced volume}}{\text{volume per shot}} = \frac{\frac{50}{3} \pi}{\frac{1}{6} \pi} = \frac{50}{3} \times 6 = 100$$ Final answers: - For 32(a): $n=7$, sum of first 20 terms = 220 - For 32(b): first term $a=2$, 13th term $a_{13}=26$ - For 33: number of lead shots = 100