Last Updated: April 2026
CBSE Class 10 Maths Chapter 5 — Arithmetic Progressions (AP) — carries ~6 marks in the Board exam (1 short + 1 long-type question typical) and is one of the most scoring chapters of the entire syllabus. The chapter introduces sequences with a constant difference, the nth term formula, and sum formulas — all of which feed directly into Class 11 Sequences & Series. For students preparing for cbse class 10 maths arithmetic progressions 2027, this complete guide covers all NCERT formulas, 7 worked problems, 20 important board-style questions and a 10-MCQ quiz aligned with the CBSE 2026-27 syllabus.
1. What is an Arithmetic Progression?
An Arithmetic Progression (AP) is a sequence in which each term differs from the previous by a fixed number called the common difference (d).
General form: a, a+d, a+2d, a+3d, …
where: a = first term; d = common difference (can be +, − or 0).
Examples:
- 2, 5, 8, 11, … (a = 2, d = 3)
- 10, 7, 4, 1, … (a = 10, d = −3)
- −5, −5, −5, … (constant AP, d = 0)
1.1 How to check if a sequence is in AP?
Check that the difference between consecutive terms is the same:
a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = d
2. nth Term of an AP
The nth (or general) term of an AP is:
aₙ = a + (n − 1)d
2.1 nth Term from the End
If l is the last term and there are n terms total:
nth term from end = l − (n − 1)d
3. Sum of First n Terms of an AP
Two equivalent forms:
S_n = (n/2)[2a + (n − 1)d]
S_n = (n/2)(a + l) (where l = aₙ = last term)
3.1 nth Term from Sum
aₙ = S_n − S_{n−1}
4. Master Formula Table (Mandatory)
| Quantity | Formula | Worked Example |
|---|---|---|
| nth term aₙ | a + (n−1)d | AP: 3, 7, 11, … 10th term = 3 + 9·4 = 39 |
| Sum of n terms S_n | (n/2)[2a + (n−1)d] | S₁₀ for above = (10/2)[6 + 36] = 210 |
| Sum (using last term) | (n/2)(a + l) | (10/2)(3 + 39) = 210 ✓ |
| Sum of first n natural nos. | n(n+1)/2 | 1+2+…+10 = 10·11/2 = 55 |
| Sum of first n squares | n(n+1)(2n+1)/6 | (beyond Class 10 syllabus, but useful) |
| Sum of first n cubes | [n(n+1)/2]² | (beyond Class 10 syllabus, but useful) |
| nth term from end | l − (n−1)d | AP 3,7,…,99: 3rd from end = 99 − 2·4 = 91 |
| aₙ from sums | S_n − S_{n−1} | If S_n = 2n²+3n: a_n = 4n+1 |
| Common difference | d = a_{n+1} − a_n | For 5, 9, 13: d = 9 − 5 = 4 |
Note: Sum of squares and cubes formulas are listed for reference but are NOT in the Class 10 CBSE 2026-27 syllabus.
5. Seven Worked Problems
Problem 1. Find the 25th term of the AP 5, 9, 13, ….
a = 5, d = 4, n = 25. a₂₅ = 5 + 24·4 = 5 + 96 = 101.
Problem 2. Which term of AP 21, 18, 15, … is −81?
a = 21, d = −3. Set 21 + (n−1)(−3) = −81 → (n−1)(−3) = −102 → n − 1 = 34 → n = 35.
Problem 3. How many two-digit numbers are divisible by 7?
14, 21, 28, …, 98. a = 14, d = 7, l = 98. l = a + (n−1)d → 98 = 14 + 7(n−1) → n − 1 = 12 → n = 13.
Problem 4. Find S₂₀ of the AP 7, 12, 17, ….
a = 7, d = 5, n = 20. S₂₀ = (20/2)[14 + 19·5] = 10[14 + 95] = 10·109 = 1090.
Problem 5. The sum of n terms of an AP is S_n = 3n² + 5n. Find the 25th term.
a₂₅ = S₂₅ − S₂₄ = (3·625 + 125) − (3·576 + 120) = 2000 − 1848 = 152.
Problem 6. Find a, given the 10th term of an AP is 21 and 18th term is 41.
a + 9d = 21; a + 17d = 41. Subtract: 8d = 20 → d = 2.5; a = 21 − 22.5 = −1.5.
Problem 7. Find the sum of all multiples of 4 between 10 and 250.
First multiple ≥ 10 is 12; last ≤ 250 is 248. AP: 12, 16, 20, …, 248. d = 4. n = (248−12)/4 + 1 = 60. S = (60/2)(12 + 248) = 30·260 = 7800.
6. 20 Important Board Questions
- Define an AP and give an example. A sequence with constant common difference. e.g., 3, 7, 11.
- Find the 11th term of AP 7, 13, 19, … = 7 + 10·6 = 67.
- Is 0 a possible common difference? Yes — gives a constant AP (all terms equal to a).
- If 4th term = 10 and 7th = 22, find d. 3d = 12 → d = 4.
- Sum of first 50 natural numbers. 50·51/2 = 1275.
- Sum of first 30 odd numbers (1, 3, 5, …). a=1, d=2, n=30. S = (30/2)(2 + 58) = 900.
- Sum of first 30 even numbers (2, 4, 6, …). S = (30/2)(2 + 60) = 930.
- Find the AP whose 3rd term is 5 and 7th term is 9. a + 2d = 5, a + 6d = 9 → d = 1, a = 3. AP: 3, 4, 5, 6, …
- Sum of first 100 positive integers divisible by 6. 6, 12, …, 600. n=100. S = (100/2)(6 + 600) = 30,300.
- How many terms are in AP 7, 13, 19, …, 205? 205 = 7 + (n−1)·6 → n = 34.
- If the sum of n terms is n² + 3n, find the AP. a₁ = S₁ = 4. a₂ = S₂ − S₁ = 10 − 4 = 6. d = 2. AP: 4, 6, 8, ….
- The 11th term from the end of AP 10, 7, 4, …, −62. l = −62, d = −3. 11th from end = −62 − 10(−3) = −62 + 30 = −32.
- Three numbers in AP have sum 27 and product 504. Find them. Let them be a−d, a, a+d. Sum = 3a = 27 → a = 9. Product: 9(81 − d²) = 504 → 81 − d² = 56 → d² = 25 → d = ±5. Numbers: 4, 9, 14 (or reverse).
- If the sum of first p terms of an AP is q and sum of first q terms is p, find the sum of (p+q) terms. S_{p+q} = −(p+q). [Standard derivation.]
- Find the AP if 2nd term = 14 and 5th term = 32. a + d = 14; a + 4d = 32 → 3d = 18 → d = 6, a = 8. AP: 8, 14, 20, …
- How many three-digit numbers are divisible by 11? 110, 121, …, 990. n = (990−110)/11 + 1 = 81.
- Find the sum of all natural numbers between 100 and 200 divisible by 4. 104, 108, …, 196. n = (196−104)/4 + 1 = 24. S = (24/2)(104 + 196) = 12·300 = 3600.
- If a, b, c are in AP, prove 2b = a + c. Definition of AP — middle term is the arithmetic mean.
- If 7 times the 7th term equals 11 times the 11th term, find the 18th term. 7(a + 6d) = 11(a + 10d) → 7a + 42d = 11a + 110d → −4a = 68d → a = −17d. a₁₈ = a + 17d = 0.
- Sum of first 20 multiples of 3. 3, 6, …, 60. n = 20. S = (20/2)(3 + 60) = 630.
7. Real-Life Applications of AP
- Saving plans (e.g., depositing ₹500 every month with an annual increase of ₹100)
- Salary increments (a fixed increment every year)
- Loan EMI structure (with constant interest difference)
- Stack of objects (logs, pipes — AP arrangements)
- Sequence of seats in a stadium row
8. Internal Resources
Build foundation with our Chapter 3 — Linear Equations and Chapter 4 — Quadratic Equations. See Class 10 Maths courses, the CBSE 2027 hub, and free resources.
9. FAQ
Q1. What is the easiest formula to remember in this chapter?
aₙ = a + (n−1)d. From this you can derive sum formulas. Most board questions can be solved with this and S_n = (n/2)(a + l).
Q2. Are sums of squares (Σn²) and cubes (Σn³) in the Class 10 syllabus?
No — those formulas are in Class 11 (Sequences & Series). Class 10 requires only AP (not GP, not Σn², not Σn³).
Q3. How can three numbers in AP be assumed for symmetric problems?
Take them as (a−d), a, (a+d). For four numbers: (a−3d), (a−d), (a+d), (a+3d) — common difference becomes 2d.
Q4. What is the marks weightage of AP in CBSE Class 10 Maths 2027?
Approximately 6 marks — typically one 2-mark VSA and one 4-mark long question, sometimes a 1-mark MCQ.
Q5. Common mistakes to avoid?
(1) Forgetting (n−1) in nth term formula; (2) using wrong sign of d; (3) confusing nth term with sum; (4) not checking if all terms in a question are in AP first; (5) arithmetic slips in S_n calculations.
Quiz — Arithmetic Progressions Class 10 — 10 MCQs
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Conclusion & CTA
AP is one of the most predictable scoring chapters — master the two key formulas (aₙ and S_n), practise 25 questions, and you’ve earned a guaranteed 6 marks. Want chapter-wise PYQ practice with step-by-step video solutions? Join Ready For Boards Class 10 Maths 2027 program.
