Arithmetic Progressions Class 10 Notes (2026-27) — CBSE
Class 10 Maths Chapter 5 notes: arithmetic progressions, the nth term formula, the sum of n terms, and word problems, with all key AP formulas.
Arithmetic Progressions — Class 10 Maths Notes
Chapter Snapshot
An arithmetic progression (AP) is a number pattern where each term increases (or decreases) by the same fixed amount. This chapter gives you two workhorse formulas — the nth term and the sum of n terms — and applies them to a wide range of problems.
Board relevance: a reliable Algebra scorer. Expect an nth-term question, a sum-of-terms question, and an AP word problem. The formulas are short and the method is mechanical once you identify a, d, and n.
Key Concepts & Definitions
Arithmetic progression (AP) — a list of numbers a₁, a₂, a₃, … in which each term is obtained by adding a fixed number to the preceding term.
Common difference (d) — the fixed number added each time: d = aₙ − aₙ₋₁. It can be positive, negative, or zero.
- General AP: a, a + d, a + 2d, a + 3d, …
- First term = a; common difference = d.
- A list is an AP only if aₖ₊₁ − aₖ is the same for all k.
Finite AP has a last term (e.g. 2, 5, 8, …, 32); infinite AP has no last term.
Formulas
The nth term (general term)
aₙ = a + (n − 1)d
- a = first term, d = common difference, n = term number, aₙ = nth term.
- nth term from the end (last term l): l − (n − 1)d.
Sum of the first n terms
Sₙ = n/2 [2a + (n − 1)d]
When the last term l = aₙ is known:
Sₙ = n/2 (a + l)
Useful relations
- d = a₂ − a₁ (check the same difference throughout).
- aₙ = Sₙ − Sₙ₋₁ (nth term from the sum).
- If a, b, c are in AP, then 2b = a + c (b is the arithmetic mean).
Worked Examples
Example 1 — Find a term: Find the 20th term of 3, 8, 13, 18, …
a = 3, d = 5, n = 20. a₂₀ = 3 + (20 − 1)(5) = 3 + 95 = 98.
Example 2 — Which term: Is 78 a term of the AP 3, 8, 13, …? If so, which?
aₙ = 3 + (n − 1)5 = 78 → 5(n − 1) = 75 → n − 1 = 15 → n = 16. Yes, the 16th term.
Example 3 — Sum: Find the sum of the first 22 terms of 8, 3, −2, …
a = 8, d = −5, n = 22. S₂₂ = 22/2 [2(8) + 21(−5)] = 11[16 − 105] = 11(−89) = −979.
Example 4 — Sum with last term: Find the sum of all two-digit numbers divisible by 3.
They are 12, 15, …, 99 → a = 12, d = 3, l = 99. First find n: 99 = 12 + (n − 1)3 → n = 30.
S₃₀ = 30/2 (12 + 99) = 15 × 111 = 1665.
Example 5 — Word problem: A sum of ₹700 is to be given as 7 cash prizes, each ₹20 less than the one before. Find each prize.
Let the prizes be an AP with sum 700, n = 7, d = −20. S₇ = 7/2[2a + 6(−20)] = 700 → 7(2a − 120) = 1400 → 2a − 120 = 200 → a = 160.
Prizes: ₹160, 140, 120, 100, 80, 60, 40.
Example 6 — Find a, d from two terms: The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.
a + 2d = 5 and a + 6d = 9. Subtracting the first from the second: 4d = 4 → d = 1, so a = 5 − 2(1) = 3.
The AP is 3, 4, 5, 6, ….
Example 7 — Rows of logs (sum): 200 logs are stacked with 20 in the bottom row, 19 in the next, and so on. How many rows hold all 200 logs?
Here a = 20, d = −1, Sₙ = 200. Sₙ = n/2[2(20) + (n − 1)(−1)] = 200 → n(41 − n) = 400 → n² − 41n + 400 = 0 → (n − 16)(n − 25) = 0.
n = 16 or n = 25; n = 25 would make the top-row count negative, so 16 rows hold the logs (a good reminder to reject the invalid value).
Important Question Patterns
1. nth term (2 marks): find a given term; find which term equals a value (solve aₙ = value for n); check if a number belongs to the AP (n must be a positive integer).
2. Sum of n terms (2–3 marks): use Sₙ = n/2[2a + (n − 1)d]; or Sₙ = n/2(a + l) when the last term is given (e.g. sum of numbers divisible by k in a range).
3. Find a, d, or n (3 marks): from two conditions (e.g. "the 4th term is 0 and the 11th term is 21"), set up two equations in a and d.
4. Given Sₙ (2–3 marks): find the nth term using aₙ = Sₙ − Sₙ₋₁; find n for a given sum.
5. Word problems (3–4 marks): savings/instalments increasing by a fixed amount, seats in rows, logs stacked, prize money — model as an AP and apply a formula.
⚡ Quick Revision
- AP: constant common difference d = aₙ − aₙ₋₁. Terms: a, a+d, a+2d, …
- nth term: aₙ = a + (n − 1)d. nth term from the end: l − (n − 1)d.
- Sum: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l) when the last term l is known.
- aₙ = Sₙ − Sₙ₋₁; if a, b, c are in AP then 2b = a + c.
- To check if a value is a term, solve aₙ = value; it belongs only if n is a positive integer.
- Identify a, d, n first; for word problems translate "increases/decreases by a fixed amount" into an AP.
Get Started Free | Features | Pricing | Blog