Given: a = 5, d = 3, a_n = 50. To find: n and S_n. Step 1: Use nth term formula: a_n = a + (n-1)d. => 50 = 5 + (n-1)×3. => 45 = (n-1)×3. => n-1 = 15. => n = 16. Step 2: Use sum formula: S_n = n/2 (a + a_n). => S_16 = 16/2 (5 + 50) = $8 \times 55$ = 440. Answer: n = 16, S_n = 440.

Given: a = 7, a_13 = 35. To find: d and S_13. Step 1: a_13 = a + 12d => 35 = 7 + 12d => 12d = 28 => d = 7/3. Step 2: S_13 = 13/2 (a + a_13) = 13/2 (7 + 35) = 13/$2 \times 42$ = $13 \times 21$ = 273. Answer: d = 7/3, S_13 = 273.

Given: a_12 = 37, d = 3. To find: a and S_12. Step 1: a_12 = a + 11d => 37 = a + 33 => a = 4. Step 2: S_12 = 12/2 (a + a_12) = 6 (4 + 37) = $6 \times 41$ = 246. Answer: a = 4, S_12 = 246.

Given: a_3 = 15, S_10 = 125. To find: d and a_10. Step 1: a_3 = a + 2d = 15. (1) Step 2: S_10 = 10/2 [2a + 9d] = 5(2a + 9d) = 125 => 2a + 9d = 25. (2) Step 3: Solve (1) and (2). From (1), a = 15 - 2d. Substitute in (2): 2(15 - 2d) + 9d = 25 => 30 - 4d + 9d = 25 => 30 + 5d = 25 => 5d = -5 => d = -1. Step 4: Then a = 15 - 2(-1) = 17. Step 5: a_10 = a + 9d = 17 + 9(-1) = 8. Answer: d = -1, a_10 = 8.

Given: d = 5, S_9 = 75. To find: a and a_9. Step 1: Use sum formula: S_9 = 9/2 [2a + 8d] = 9/2 (2a + 40) = 9(a + 20) (since 9/$2 \times 2$(a+20) = 9(a+20)). => 9(a + 20) = 75 => a + 20 = 75/9 = 25/3 => a = 25/3 - 20 = -35/3. Step 2: a_9 = a + 8d = -35/3 + 40 = -35/3 + 120/3 = 85/3. Answer: a = -35/3, a_9 = 85/3.

Given: a = 2, d = 8, S_n = 90. To find: n and a_n. Step 1: S_n = n/2 [2a + (n-1)d] = n/2 [4 + (n-1)×8] = n/2 (8n - 4) = n/$2 \times 4$(2n-1) = 2n(2n-1). => 2n(2n-1) = 90 => n(2n-1) = 45 => $2n^{2}$ - n - 45 = 0. Step 2: Solve quadratic: $2n^{2}$ - n - 45 = 0. Discriminant = 1 + 360 = 361, sqrt = 19. => n = [1 ± 19] / 4. Positive root: n = (1+19)/4 = 5. Step 3: a_n = a + (n-1)d = 2 + $4 \times 8$ = 34. Answer: n = 5, a_n = 34.

Given: a = 8, a_n = 62, S_n = 210. To find: n and d. Step 1: S_n = n/2 (a + a_n) => 210 = n/2 (8+62) = n/$2 \times 70$ = 35n => n = 210/35 = 6. Step 2: a_n = a + (n-1)d => 62 = 8 + 5d => 5d = 54 => d = 54/5. Answer: n = 6, d = 54/5.

Given: a_n = 4, d = 2, S_n = -14. To find: n and a. Step 1: a_n = a + (n-1)d => 4 = a + 2(n-1) => a = 4 - 2(n-1) = 6 - 2n. Step 2: S_n = n/2 (a + a_n) = n/2 (a + 4). Substitute a: n/2 (6 - 2n + 4) = n/2 (10 - 2n) = n(5 - n) = -14. => n(5 - n) = -14 => 5n - $n^{2}$ = -14 => $n^{2}$ - 5n - 14 = 0. Step 3: Solve: $n^{2}$ - 5n - 14 = 0 => (n-7)(n+2)=0 => n = 7 (since n>0). Step 4: a = 6 - 2n = 6 - 14 = -8. Answer: n = 7, a = -8.

Given: a = 3, n = 8, S = 192. To find: d. Step 1: S = n/2 [2a + (n-1)d] => 192 = 8/2 [6 + 7d] = 4(6 + 7d) => 6 + 7d = 192/4 = 48 => 7d = 42 => d = 6. Answer: d = 6.

Given: l = 28, S = 144, n = 9. To find: a. Step 1: Use sum formula: S = n/2 (a + l) => 144 = 9/2 (a + 28) => Multiply by 2: 288 = 9(a + 28) => a + 28 = 288/9 = 32 => a = 32 - 28 = 4. Answer: a = 4.