CBSE Class 10 Maths: Arithmetic Progressions — Notes 2026

Introduction to AP

• Sequence where each term differs from previous by fixed amount called common difference (d)
• General form: a, a+d, a+2d, a+3d, … ; first term = a, common difference = d
• Example: 2, 5, 8, 11, 14 … → a = 2, d = 3; example with negative d: 20, 17, 14, 11 → d = −3

nth Term Formula

• aₙ = a + (n−1)d; also written as l (last term) when dealing with finite AP
• Find a₁₅ of AP 3, 7, 11, 15 … : a=3, d=4; a₁₅ = 3 + 14×4 = 59
• Find n when aₙ = given value: solve (a + (n−1)d = given) for n; ensure n is positive integer

Sum of n Terms

• Sₙ = n/2 × (2a + (n−1)d) = n/2 × (a + l) where l is last term
• Example: sum of first 20 terms of AP 5, 8, 11…: S₂₀ = 20/2 × (10 + 19×3) = 10 × 67 = 670
• Derivation: write sum forwards and backwards; add both; each pair = a + l; n pairs → Sₙ = n(a+l)/2

Finding AP from Given Conditions

• If 3rd term = 7 and 7th term = 19: a+2d = 7 and a+6d = 19 → solve simultaneously → a=4, d=3
• Given sum S and last term l: use Sₙ = n/2(a+l) with n = (l−a)/d + 1
• Given Sₙ = pn² + qn: aₙ = Sₙ − Sₙ₋₁ = 2pn + (q−p)

Middle Term and Arithmetic Mean

• Arithmetic mean of a and b: A = (a+b)/2; insert n arithmetic means between a and b: d = (b−a)/(n+1)
• For 3 terms in AP: take as a−d, a, a+d → sum = 3a; product involves a and d
• CBSE trick: sum of 5 terms in AP = 5×middle term; sum of 3 terms = 3×middle term

Word Problems

• Salary AP: annual increment problems; compound interest (no, that's GP) — beware confusion
• Seating arrangement: theatre with n rows, 1st row = x seats, each row increases by y seats; find total seats
• Penalty/bonus: contractor fined increasing amount each day → AP sum = total fine

CBSE Exam Strategy

• AP is 5–8 marks in CBSE; both short and long answer questions common
• Always state a, d, and n clearly before using formula; show substitution step
• If Sₙ given, find aₙ by aₙ = Sₙ − Sₙ₋₁; this formula often tested