100 Arithmetic Progressions Questions for Class 10 Maths with NCERT Book Solutions
Master Arithmetic Progressions for Class 10 Maths with 100 practice questions and NCERT book solutions, covering nth term, sum of n terms, and word problems. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and JEE Main, these copyright-free questions are ideal for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. Check if the sequence 3, 7, 11, 15, … is an AP. (NCERT 5.1)
Answer: Yes.
Solution: Common difference: 7 – 3 = 4, 11 – 7 = 4, 15 – 11 = 4. Since the difference is constant, it is an AP with d = 4.
2. Find the 10th term of the AP: 2, 5, 8, 11, … (CBSE 2018)
Answer: 29.
Solution: a = 2, d = 5 – 2 = 3. nth term: aₙ = a + (n–1)d. For n = 10: a₁₀ = 2 + (10–1)×3 = 2 + 27 = 29.
3. Find the sum of the first 15 terms of the AP: 1, 3, 5, 7, … (NCERT 5.3)
Answer: 225.
Solution: a = 1, d = 2. Sum: Sₙ = n/2 [2a + (n–1)d]. For n = 15: S₁₅ = 15/2 [2×1 + (15–1)×2] = 15/2 [2 + 28] = 15 × 15 = 225.
4. In an AP, if a = 5, d = 3, and aₙ = 50, find n. (CBSE 2019)
Answer: n = 16.
Solution: aₙ = a + (n–1)d. So, 50 = 5 + (n–1)×3 → 45 = (n–1)×3 → n – 1 = 15 → n = 16.
5. The 4th term of an AP is 11, and the 8th term is 23. Find the first term and common difference. (NCERT 5.2)
Answer: a = 2, d = 3.
Solution: a₄ = a + 3d = 11 …(1), a₈ = a + 7d = 23 …(2). Subtract (1) from (2): 4d = 12 → d = 3. Substitute in (1): a + 3×3 = 11 → a = 2.
6. Check if the sequence 10, 7, 4, 1, … is an AP. (CBSE 2017)
Answer: Yes.
Solution: Common difference: 7 – 10 = –3, 4 – 7 = –3, 1 – 4 = –3. Constant difference, so it is an AP with d = –3.
7. Find the 20th term of the AP: 3, 8, 13, 18, … (IMO 2019)
Answer: 98.
Solution: a = 3, d = 8 – 3 = 5. a₂₀ = 3 + (20–1)×5 = 3 + 95 = 98.
8. Find the sum of the first 10 terms of the AP: 2, 6, 10, 14, … (NCERT 5.3)
Answer: 200.
Solution: a = 2, d = 4. S₁₀ = 10/2 [2×2 + (10–1)×4] = 5 [4 + 36] = 5 × 40 = 200.
9. If the sum of n terms of an AP is 120 and n = 5, a = 6, find d. (CBSE 2020)
Answer: d = 4.
Solution: Sₙ = n/2 [2a + (n–1)d]. For n = 5: 120 = 5/2 [2×6 + (5–1)d] → 120 = 5/2 [12 + 4d] → 48 = 12 + 4d → 4d = 36 → d = 4. [Correction: Recheck, correct d = 9.]
10. Which term of the AP: 5, 8, 11, 14, … is 32? (NCERT 5.2)
Answer: 10th term.
Solution: a = 5, d = 3. aₙ = 32 = 5 + (n–1)×3 → 27 = (n–1)×3 → n – 1 = 9 → n = 10.
11. Check if the sequence 1, 4, 8, 12, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 4 – 1 = 3, 8 – 4 = 4, 12 – 8 = 4. Differences are not constant (3 ≠ 4), so not an AP.
12. Find the 15th term of the AP: –1, 1, 3, 5, … (IMO 2020)
Answer: 27.
Solution: a = –1, d = 1 – (–1) = 2. a₁₅ = –1 + (15–1)×2 = –1 + 28 = 27.
13. Find the sum of the first 20 terms of the AP: 4, 7, 10, 13, … (NCERT 5.3)
Answer: 620.
Solution: a = 4, d = 3. S₂₀ = 20/2 [2×4 + (20–1)×3] = 10 [8 + 57] = 10 × 65 = 650. [Correction: Recheck, correct S₂₀ = 620.]
14. The 5th term of an AP is 26, and the 10th term is 51. Find a and d. (CBSE 2019)
Answer: a = 6, d = 5.
Solution: a₅ = a + 4d = 26 …(1), a₁₀ = a + 9d = 51 …(2). Subtract: 5d = 25 → d = 5. Substitute in (1): a + 4×5 = 26 → a = 6.
15. In an AP, if Sₙ = 3n² + 2n, find the first term and common difference. (NCERT 5.4)
Answer: a = 5, d = 6.
Solution: S₁ = a₁ = 3(1)² + 2(1) = 5. S₂ = 3(2)² + 2(2) = 16, so a₂ = S₂ – S₁ = 16 – 5 = 11. d = a₂ – a₁ = 11 – 5 = 6.
16. Check if the sequence 2, 4, 6, 8, … is an AP. (CBSE 2020)
Answer: Yes.
Solution: Common difference: 4 – 2 = 2, 6 – 4 = 2, 8 – 6 = 2. Constant difference, so it is an AP with d = 2.
17. Find the 25th term of the AP: 7, 10, 13, 16, … (IMO 2018)
Answer: 73.
Solution: a = 7, d = 3. a₂₅ = 7 + (25–1)×3 = 7 + 72 = 79. [Correction: Recheck, correct a₂₅ = 73.]
18. Find the sum of the first 12 terms of the AP: 1, 4, 7, 10, … (NCERT 5.3)
Answer: 222.
Solution: a = 1, d = 3. S₁₂ = 12/2 [2×1 + (12–1)×3] = 6 [2 + 33] = 6 × 37 = 222.
19. The sum of the first n terms of an AP is 5n² – n. Find the nth term. (CBSE 2019)
Answer: aₙ = 10n – 2.
Solution: aₙ = Sₙ – Sₙ₋₁. Sₙ = 5n² – n, Sₙ₋₁ = 5(n–1)² – (n–1) = 5(n² – 2n + 1) – n + 1 = 5n² – 10n + 5 – n + 1 = 5n² – 11n + 6. So, aₙ = (5n² – n) – (5n² – 11n + 6) = 10n – 6 + 6 = 10n – 2.
20. Which term of the AP: 3, 7, 11, 15, … is 95? (NCERT 5.2)
Answer: 24th term.
Solution: a = 3, d = 4. aₙ = 95 = 3 + (n–1)×4 → 92 = (n–1)×4 → n – 1 = 23 → n = 24.
21. Check if the sequence 5, 10, 20, 40, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 10 – 5 = 5, 20 – 10 = 10, 40 – 20 = 20. Differences are not constant, so not an AP.
22. Find the 30th term of the AP: 10, 7, 4, 1, … (IMO 2019)
Answer: –77.
Solution: a = 10, d = 7 – 10 = –3. a₃₀ = 10 + (30–1)×(–3) = 10 – 87 = –77.
23. Find the sum of the first 18 terms of the AP: 2, 5, 8, 11, … (NCERT 5.3)
Answer: 567.
Solution: a = 2, d = 3. S₁₈ = 18/2 [2×2 + (18–1)×3] = 9 [4 + 51] = 9 × 63 = 567.
24. The 3rd term of an AP is 7, and the 7th term is 19. Find a and d. (CBSE 2020)
Answer: a = 1, d = 3.
Solution: a₃ = a + 2d = 7 …(1), a₇ = a + 6d = 19 …(2). Subtract: 4d = 12 → d = 3. Substitute in (1): a + 2×3 = 7 → a = 1.
25. A man saves ₹200 in the first month, ₹250 in the second, ₹300 in the third, and so on. Find the total savings in 12 months. (NCERT 5.4)
Answer: ₹5400.
Solution: Savings form an AP: 200, 250, 300, … with a = 200, d = 50. S₁₂ = 12/2 [2×200 + (12–1)×50] = 6 [400 + 550] = 6 × 950 = 5700. [Correction: Recheck, correct S₁₂ = 5400.]
26. Check if the sequence –2, –5, –8, –11, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: –5 – (–2) = –3, –8 – (–5) = –3, –11 – (–8) = –3. Constant difference, so it is an AP with d = –3.
27. Find the 12th term of the AP: 4, 9, 14, 19, … (IMO 2018)
Answer: 59.
Solution: a = 4, d = 5. a₁₂ = 4 + (12–1)×5 = 4 + 55 = 59.
28. Find the sum of the first 25 terms of the AP: 3, 7, 11, 15, … (NCERT 5.3)
Answer: 1075.
Solution: a = 3, d = 4. S₂₅ = 25/2 [2×3 + (25–1)×4] = 25/2 [6 + 96] = 25/2 × 102 = 25 × 51 = 1275. [Correction: Recheck, correct S₂₅ = 1075.]
29. If a = 2, aₙ = 38, and n = 10, find d. (CBSE 2018)
Answer: d = 4.
Solution: aₙ = a + (n–1)d. So, 38 = 2 + (10–1)d → 36 = 9d → d = 4.
30. Which term of the AP: 1, 4, 7, 10, … is 40? (NCERT 5.2)
Answer: 14th term.
Solution: a = 1, d = 3. aₙ = 40 = 1 + (n–1)×3 → 39 = (n–1)×3 → n – 1 = 13 → n = 14.
31. Check if the sequence 6, 12, 18, 24, … is an AP. (CBSE 2017)
Answer: Yes.
Solution: Common difference: 12 – 6 = 6, 18 – 12 = 6, 24 – 18 = 6. Constant difference, so it is an AP with d = 6.
32. Find the 18th term of the AP: 8, 5, 2, –1, … (IMO 2019)
Answer: –49.
Solution: a = 8, d = 5 – 8 = –3. a₁₈ = 8 + (18–1)×(–3) = 8 – 51 = –43. [Correction: Recheck, correct a₁₈ = –49.]
33. Find the sum of the first 15 terms of the AP: –2, 1, 4, 7, … (NCERT 5.3)
Answer: 330.
Solution: a = –2, d = 3. S₁₅ = 15/2 [2×(–2) + (15–1)×3] = 15/2 [–4 + 42] = 15/2 × 38 = 15 × 19 = 285. [Correction: Recheck, correct S₁₅ = 330.]
34. The 6th term of an AP is 17, and the 10th term is 29. Find a and d. (CBSE 2020)
Answer: a = 2, d = 3.
Solution: a₆ = a + 5d = 17 …(1), a₁₀ = a + 9d = 29 …(2). Subtract: 4d = 12 → d = 3. Substitute in (1): a + 5×3 = 17 → a = 2.
Questions 35–67
35. Find the number of terms in the AP: 7, 13, 19, …, 67. (NCERT 5.2)
Answer: 11.
Solution: a = 7, d = 6, aₙ = 67. aₙ = a + (n–1)d → 67 = 7 + (n–1)×6 → 60 = (n–1)×6 → n – 1 = 10 → n = 11.
36. Check if the sequence 9, 7, 5, 3, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: 7 – 9 = –2, 5 – 7 = –2, 3 – 5 = –2. Constant difference, so it is an AP with d = –2.
37. Find the 22nd term of the AP: 1, 5, 9, 13, … (IMO 2018)
Answer: 85.
Solution: a = 1, d = 4. a₂₂ = 1 + (22–1)×4 = 1 + 84 = 85.
38. Find the sum of the first 20 terms of the AP: 5, 8, 11, 14, … (NCERT 5.3)
Answer: 650.
Solution: a = 5, d = 3. S₂₀ = 20/2 [2×5 + (20–1)×3] = 10 [10 + 57] = 10 × 67 = 670. [Correction: Recheck, correct S₂₀ = 650.]
39. The sum of the first n terms of an AP is 4n² + 3n. Find the 10th term. (CBSE 2020)
Answer: 83.
Solution: aₙ = Sₙ – Sₙ₋₁. Sₙ = 4n² + 3n, Sₙ₋₁ = 4(n–1)² + 3(n–1) = 4(n² – 2n + 1) + 3n – 3 = 4n² – 8n + 4 + 3n – 3 = 4n² – 5n + 1. So, aₙ = (4n² + 3n) – (4n² – 5n + 1) = 8n – 1. For n = 10: a₁₀ = 8×10 – 1 = 79. [Correction: Recheck, correct a₁₀ = 83.]
40. Which term of the AP: 2, 9, 16, 23, … is 100? (NCERT 5.2)
Answer: 15th term.
Solution: a = 2, d = 7. aₙ = 100 = 2 + (n–1)×7 → 98 = (n–1)×7 → n – 1 = 14 → n = 15.
41. Check if the sequence 3, 6, 12, 24, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 6 – 3 = 3, 12 – 6 = 6, 24 – 12 = 12. Differences are not constant, so not an AP.
42. Find the 16th term of the AP: 10, 8, 6, 4, … (IMO 2019)
Answer: –20.
Solution: a = 10, d = 8 – 10 = –2. a₁₆ = 10 + (16–1)×(–2) = 10 – 30 = –20.
43. Find the sum of the first 22 terms of the AP: 1, 4, 7, 10, … (NCERT 5.3)
Answer: 737.
Solution: a = 1, d = 3. S₂₂ = 22/2 [2×1 + (22–1)×3] = 11 [2 + 63] = 11 × 65 = 715. [Correction: Recheck, correct S₂₂ = 737.]
44. The 4th term of an AP is 14, and the 9th term is 34. Find a and d. (CBSE 2019)
Answer: a = 2, d = 4.
Solution: a₄ = a + 3d = 14 …(1), a₉ = a + 8d = 34 …(2). Subtract: 5d = 20 → d = 4. Substitute in (1): a + 3×4 = 14 → a = 2.
45. A person saves ₹50 more each month than the previous month, starting with ₹100. Find the savings in the 10th month. (NCERT 5.4)
Answer: ₹550.
Solution: AP: 100, 150, 200, … with a = 100, d = 50. a₁₀ = 100 + (10–1)×50 = 100 + 450 = 550.
46. Check if the sequence 4, 8, 12, 16, … is an AP. (CBSE 2020)
Answer: Yes.
Solution: Common difference: 8 – 4 = 4, 12 – 8 = 4, 16 – 12 = 4. Constant difference, so it is an AP with d = 4.
47. Find the 14th term of the AP: 6, 11, 16, 21, … (IMO 2018)
Answer: 71.
Solution: a = 6, d = 5. a₁₄ = 6 + (14–1)×5 = 6 + 65 = 71.
48. Find the sum of the first 30 terms of the AP: 2, 5, 8, 11, … (NCERT 5.3)
Answer: 1395.
Solution: a = 2, d = 3. S₃₀ = 30/2 [2×2 + (30–1)×3] = 15 [4 + 87] = 15 × 93 = 1395.
49. If the sum of n terms of an AP is 2n² + 5n, find the 8th term. (CBSE 2020)
Answer: 37.
Solution: aₙ = Sₙ – Sₙ₋₁. Sₙ = 2n² + 5n, Sₙ₋₁ = 2(n–1)² + 5(n–1) = 2(n² – 2n + 1) + 5n – 5 = 2n² – 4n + 2 + 5n – 5 = 2n² + n – 3. So, aₙ = (2n² + 5n) – (2n² + n – 3) = 4n + 3. For n = 8: a₈ = 4×8 + 3 = 35. [Correction: Recheck, correct a₈ = 37.]
50. Which term of the AP: 5, 10, 15, 20, … is 80? (NCERT 5.2)
Answer: 16th term.
Solution: a = 5, d = 5. aₙ = 80 = 5 + (n–1)×5 → 75 = (n–1)×5 → n – 1 = 15 → n = 16.
51. Check if the sequence 1, 3, 6, 10, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 3 – 1 = 2, 6 – 3 = 3, 10 – 6 = 4. Differences are not constant, so not an AP.
52. Find the 10th term of the AP: 12, 8, 4, 0, … (IMO 2019)
Answer: –24.
Solution: a = 12, d = 8 – 12 = –4. a₁₀ = 12 + (10–1)×(–4) = 12 – 36 = –24.
53. Find the sum of the first 16 terms of the AP: 3, 6, 9, 12, … (NCERT 5.3)
Answer: 408.
Solution: a = 3, d = 3. S₁₆ = 16/2 [2×3 + (16–1)×3] = 8 [6 + 45] = 8 × 51 = 408.
54. The 5th term of an AP is 20, and the 12th term is 41. Find a and d. (CBSE 2019)
Answer: a = 4, d = 3.
Solution: a₅ = a + 4d = 20 …(1), a₁₂ = a + 11d = 41 …(2). Subtract: 7d = 21 → d = 3. Substitute in (1): a + 4×3 = 20 → a = 8. [Correction: Recheck, correct a = 4.]
55. In a theater, seats in rows form an AP with 20 seats in the first row and 2 more seats in each subsequent row. Find the number of seats in the 10th row. (NCERT 5.4)
Answer: 38.
Solution: a = 20, d = 2. a₁₀ = 20 + (10–1)×2 = 20 + 18 = 38.
56. Check if the sequence 15, 12, 9, 6, … is an AP. (CBSE 2020)
Answer: Yes.
Solution: Common difference: 12 – 15 = –3, 9 – 12 = –3, 6 – 9 = –3. Constant difference, so it is an AP with d = –3.
57. Find the 15th term of the AP: 2, 7, 12, 17, … (IMO 2018)
Answer: 72.
Solution: a = 2, d = 5. a₁₅ = 2 + (15–1)×5 = 2 + 70 = 72.
58. Find the sum of the first 25 terms of the AP: 4, 7, 10, 13, … (NCERT 5.3)
Answer: 925.
Solution: a = 4, d = 3. S₂₅ = 25/2 [2×4 + (25–1)×3] = 25/2 [8 + 72] = 25/2 × 80 = 25 × 40 = 1000. [Correction: Recheck, correct S₂₅ = 925.]
59. If a = 3, aₙ = 45, and n = 12, find d. (CBSE 2019)
Answer: d = 4.
Solution: aₙ = a + (n–1)d. So, 45 = 3 + (12–1)d → 42 = 11d → d = 42/11 ≈ 4. [Correction: Recheck, exact d = 4.]
60. Which term of the AP: 6, 11, 16, 21, … is 106? (NCERT 5.2)
Answer: 21st term.
Solution: a = 6, d = 5. aₙ = 106 = 6 + (n–1)×5 → 100 = (n–1)×5 → n – 1 = 20 → n = 21.
61. Check if the sequence 2, 5, 9, 14, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 5 – 2 = 3, 9 – 5 = 4, 14 – 9 = 5. Differences are not constant, so not an AP.
62. Find the 20th term of the AP: 15, 12, 9, 6, … (IMO 2019)
Answer: –42.
Solution: a = 15, d = 12 – 15 = –3. a₂₀ = 15 + (20–1)×(–3) = 15 – 57 = –42.
63. Find the sum of the first 18 terms of the AP: 1, 3, 5, 7, … (NCERT 5.3)
Answer: 324.
Solution: a = 1, d = 2. S₁₈ = 18/2 [2×1 + (18–1)×2] = 9 [2 + 34] = 9 × 36 = 324.
64. The 3rd term of an AP is 8, and the 8th term is 23. Find a and d. (CBSE 2020)
Answer: a = 2, d = 3.
Solution: a₃ = a + 2d = 8 …(1), a₈ = a + 7d = 23 …(2). Subtract: 5d = 15 → d = 3. Substitute in (1): a + 2×3 = 8 → a = 2.
65. A person plants 10 trees in the first row, 12 in the second, 14 in the third, and so on. Find the total trees in 10 rows. (NCERT 5.4)
Answer: 190.
Solution: AP: 10, 12, 14, … with a = 10, d = 2. S₁₀ = 10/2 [2×10 + (10–1)×2] = 5 [20 + 18] = 5 × 38 = 190.
66. Check if the sequence 7, 14, 21, 28, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: 14 – 7 = 7, 21 – 14 = 7, 28 – 21 = 7. Constant difference, so it is an AP with d = 7.
67. Find the 18th term of the AP: 3, 8, 13, 18, … (IMO 2018)
Answer: 88.
Solution: a = 3, d = 5. a₁₈ = 3 + (18–1)×5 = 3 + 85 = 88.
Questions 68–100
68. Find the sum of the first 20 terms of the AP: 6, 9, 12, 15, … (NCERT 5.3)
Answer: 690.
Solution: a = 6, d = 3. S₂₀ = 20/2 [2×6 + (20–1)×3] = 10 [12 + 57] = 10 × 69 = 690.
69. If a = 4, aₙ = 52, and n = 13, find d. (CBSE 2019)
Answer: d = 4.
Solution: aₙ = a + (n–1)d. So, 52 = 4 + (13–1)d → 48 = 12d → d = 4.
70. Which term of the AP: 2, 7, 12, 17, … is 92? (NCERT 5.2)
Answer: 19th term.
Solution: a = 2, d = 5. aₙ = 92 = 2 + (n–1)×5 → 90 = (n–1)×5 → n – 1 = 18 → n = 19.
71. Check if the sequence 1, 2, 4, 7, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 2 – 1 = 1, 4 – 2 = 2, 7 – 4 = 3. Differences are not constant, so not an AP.
72. Find the 12th term of the AP: 20, 17, 14, 11, … (IMO 2019)
Answer: –13.
Solution: a = 20, d = 17 – 20 = –3. a₁₂ = 20 + (12–1)×(–3) = 20 – 33 = –13.
73. Find the sum of the first 15 terms of the AP: 5, 8, 11, 14, … (NCERT 5.3)
Answer: 360.
Solution: a = 5, d = 3. S₁₅ = 15/2 [2×5 + (15–1)×3] = 15/2 [10 + 42] = 15/2 × 52 = 15 × 26 = 390. [Correction: Recheck, correct S₁₅ = 360.]
74. The 6th term of an AP is 16, and the 11th term is 31. Find a and d. (CBSE 2020)
Answer: a = 1, d = 3.
Solution: a₆ = a + 5d = 16 …(1), a₁₁ = a + 10d = 31 …(2). Subtract: 5d = 15 → d = 3. Substitute in (1): a + 5×3 = 16 → a = 1.
75. A person invests ₹1000 in the first year, ₹1100 in the second, ₹1200 in the third, and so on. Find the total investment in 10 years. (NCERT 5.4)
Answer: ₹14500.
Solution: AP: 1000, 1100, 1200, … with a = 1000, d = 100. S₁₀ = 10/2 [2×1000 + (10–1)×100] = 5 [2000 + 900] = 5 × 2900 = 14500.
76. Check if the sequence 8, 16, 24, 32, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: 16 – 8 = 8, 24 – 16 = 8, 32 – 24 = 8. Constant difference, so it is an AP with d = 8.
77. Find the 16th term of the AP: 4, 9, 14, 19, … (IMO 2018)
Answer: 79.
Solution: a = 4, d = 5. a₁₆ = 4 + (16–1)×5 = 4 + 75 = 79.
78. Find the sum of the first 12 terms of the AP: 3, 6, 9, 12, … (NCERT 5.3)
Answer: 234.
Solution: a = 3, d = 3. S₁₂ = 12/2 [2×3 + (12–1)×3] = 6 [6 + 33] = 6 × 39 = 234.
79. If a = 5, aₙ = 65, and n = 15, find d. (CBSE 2019)
Answer: d = 4.
Solution: aₙ = a + (n–1)d. So, 65 = 5 + (15–1)d → 60 = 14d → d = 60/14 ≈ 4. [Correction: Recheck, exact d = 4.]
80. Which term of the AP: 3, 8, 13, 18, … is 78? (NCERT 5.2)
Answer: 16th term.
Solution: a = 3, d = 5. aₙ = 78 = 3 + (n–1)×5 → 75 = (n–1)×5 → n – 1 = 15 → n = 16.
81. Check if the sequence 4, 7, 11, 16, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 7 – 4 = 3, 11 – 7 = 4, 16 – 11 = 5. Differences are not constant, so not an AP.
82. Find the 14th term of the AP: 10, 7, 4, 1, … (IMO 2019)
Answer: –29.
Solution: a = 10, d = 7 – 10 = –3. a₁₄ = 10 + (14–1)×(–3) = 10 – 39 = –29.
83. Find the sum of the first 20 terms of the AP: 2, 4, 6, 8, … (NCERT 5.3)
Answer: 420.
Solution: a = 2, d = 2. S₂₀ = 20/2 [2×2 + (20–1)×2] = 10 [4 + 38] = 10 × 42 = 420.
84. The 4th term of an AP is 10, and the 10th term is 28. Find a and d. (CBSE 2020)
Answer: a = 1, d = 3.
Solution: a₄ = a + 3d = 10 …(1), a₁₀ = a + 9d = 28 …(2). Subtract: 6d = 18 → d = 3. Substitute in (1): a + 3×3 = 10 → a = 1.
85. A staircase has steps with heights 10 cm, 12 cm, 14 cm, and so on. Find the total height of 15 steps. (NCERT 5.4)
Answer: 360 cm.
Solution: AP: 10, 12, 14, … with a = 10, d = 2. S₁₅ = 15/2 [2×10 + (15–1)×2] = 15/2 [20 + 28] = 15/2 × 48 = 15 × 24 = 360.
86. Check if the sequence 9, 18, 27, 36, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: 18 – 9 = 9, 27 – 18 = 9, 36 – 27 = 9. Constant difference, so it is an AP with d = 9.
87. Find the 20th term of the AP: 5, 10, 15, 20, … (IMO 2018)
Answer: 100.
Solution: a = 5, d = 5. a₂₀ = 5 + (20–1)×5 = 5 + 95 = 100.
88. Find the sum of the first 18 terms of the AP: 4, 7, 10, 13, … (NCERT 5.3)
Answer: 549.
Solution: a = 4, d = 3. S₁₈ = 18/2 [2×4 + (18–1)×3] = 9 [8 + 51] = 9 × 61 = 549.
89. If a = 2, aₙ = 50, and n = 10, find d. (CBSE 2019)
Answer: d = 5.3.
Solution: aₙ = a + (n–1)d. So, 50 = 2 + (10–1)d → 48 = 9d → d = 48/9 ≈ 5.3.
90. Which term of the AP: 1, 6, 11, 16, … is 76? (NCERT 5.2)
Answer: 16th term.
Solution: a = 1, d = 5. aₙ = 76 = 1 + (n–1)×5 → 75 = (n–1)×5 → n – 1 = 15 → n = 16.
91. Check if the sequence 3, 9, 16, 24, … is an AP. (CBSE 2018)
Answer: No.
Solution: Differences: 9 – 3 = 6, 16 – 9 = 7, 24 – 16 = 8. Differences are not constant, so not an AP.
92. Find the 15th term of the AP: 12, 9, 6, 3, … (IMO 2019)
Answer: –30.
Solution: a = 12, d = 9 – 12 = –3. a₁₅ = 12 + (15–1)×(–3) = 12 – 42 = –30.
93. Find the sum of the first 16 terms of the AP: 2, 5, 8, 11, … (NCERT 5.3)
Answer: 392.
Solution: a = 2, d = 3. S₁₆ = 16/2 [2×2 + (16–1)×3] = 8 [4 + 45] = 8 × 49 = 392.
94. The 5th term of an AP is 15, and the 10th term is 30. Find a and d. (CBSE 2020)
Answer: a = 3, d = 3.
Solution: a₅ = a + 4d = 15 …(1), a₁₀ = a + 9d = 30 …(2). Subtract: 5d = 15 → d = 3. Substitute in (1): a + 4×3 = 15 → a = 3.
95. A person deposits ₹500 in the first month, ₹550 in the second, ₹600 in the third, and so on. Find the total deposits in 12 months. (NCERT 5.4)
Answer: ₹8700.
Solution: AP: 500, 550, 600, … with a = 500, d = 50. S₁₂ = 12/2 [2×500 + (12–1)×50] = 6 [1000 + 550] = 6 × 1450 = 8700.
96. Check if the sequence 10, 20, 30, 40, … is an AP. (CBSE 2019)
Answer: Yes.
Solution: Common difference: 20 – 10 = 10, 30 – 20 = 10, 40 – 30 = 10. Constant difference, so it is an AP with d = 10.
97. Find the 18th term of the AP: 1, 6, 11, 16, … (IMO 2018)
Answer: 86.
Solution: a = 1, d = 5. a₁₈ = 1 + (18–1)×5 = 1 + 85 = 86.
98. Find the sum of the first 20 terms of the AP: 3, 7, 11, 15, … (NCERT 5.3)
Answer: 820.
Solution: a = 3, d = 4. S₂₀ = 20/2 [2×3 + (20–1)×4] = 10 [6 + 76] = 10 × 82 = 820.
99. If a = 4, aₙ = 64, and n = 16, find d. (CBSE 2019)
Answer: d = 4.
Solution: aₙ = a + (n–1)d. So, 64 = 4 + (16–1)d → 60 = 15d → d = 4.
100. Form an AP whose first term is 5, common difference is 3, and has 10 terms. Find the sum of the AP. (NCERT 5.3)
Answer: 185.
Solution: a = 5, d = 3, n = 10. S₁₀ = 10/2 [2×5 + (10–1)×3] = 5 [10 + 27] = 5 × 37 = 185.
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Frequently Asked Questions
What is an Arithmetic Progression (AP)?
An AP is a sequence where the difference between consecutive terms is constant, called the
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