1. Problem 32(a): An arithmetic progression (AP) has $n$ terms with the $n$th term $a_n = 4$ and 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. Express $a$ in terms of $n$: $$a = 4 - 2(n-1) = 4 - 2n + 2 = 6 - 2n$$ 6. Substitute $a$ into the sum formula and set $S_n = -14$: $$-14 = \frac{n}{2}[2(6 - 2n) + (n-1)2]$$ 7. Simplify inside the bracket: $$2(6 - 2n) + 2(n-1) = 12 - 4n + 2n - 2 = 10 - 2n$$ 8. So, $$-14 = \frac{n}{2}(10 - 2n) = n(5 - n)$$ 9. Rearranged: $$n(5 - n) = -14 \implies 5n - n^2 = -14 \implies n^2 - 5n - 14 = 0$$ 10. Solve quadratic: $$n = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2}$$ 11. Possible $n$ values: $$n = 7 \text{ or } n = -2$$ (discard negative) 12. So, $n = 7$. 13. Find $a$: $$a = 6 - 2(7) = 6 - 14 = -8$$ 14. Find sum of first 20 terms: $$S_{20} = \frac{20}{2}[2(-8) + (20-1)2] = 10[-16 + 38] = 10 \times 22 = 220$$ **Final answers:** - Number of terms $n = 7$ - Sum of first 20 terms $S_{20} = 220$