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Class 10 Maths Notes — Chapter 5: Arithmetic Progressions

Sequences with a constant difference between successive terms — general term and sum formulas.

Detailed NCERT notes

  • AP = list of numbers a, a+d, a+2d, a+3d, … in which the difference between any two consecutive terms is a constant d, called the common difference.
  • First term = a, common difference d = aₙ₊₁ − aₙ (should be the same for all n).
  • General (nth) term: aₙ = a + (n − 1)d.
  • Sum of first n terms: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l), where l = last term.
  • If Sₙ is given as a polynomial in n, then aₙ = Sₙ − Sₙ₋₁.
  • To find how many terms: use aₙ formula and equate to the last term; n must be a positive integer.
  • For 3 numbers in AP whose sum is known, take them as (a − d, a, a + d) — sum simplifies to 3a.
  • For 4 numbers in AP with a symmetric condition, take (a − 3d, a − d, a + d, a + 3d) — sum = 4a, common difference = 2d.
  • Middle term of an AP with odd number of terms lies at position (n+1)/2.
  • Applications: instalment problems, seating arrangements, salary increments, stack/pyramid counting.

Formulas & key results

  • aₙ = a + (n − 1)d
  • Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
  • aₙ = Sₙ − Sₙ₋₁

Mind map

  • AP → common difference d constant
  • nth term → sum of n terms
  • 3 or 4 unknowns → symmetric substitution

Tricks & shortcuts

  • 3 unknowns in AP: take a−d, a, a+d.
  • 4 unknowns in AP: take a−3d, a−d, a+d, a+3d.
  • Sum-based problems: prefer Sₙ = n/2 (a + l) if last term is known.

Common mistakes to avoid

  • Using n instead of n − 1 in aₙ.
  • Confusing last term l with number of terms n.
  • Assuming a list is AP without checking constant d.

Competency-based questions & answers

  1. Q. Sum of first 20 terms of AP: 2, 7, 12, …
    A. a = 2, d = 5, S₂₀ = 20/2 (4 + 19·5) = 10·99 = 990.
  2. Q. Which term of the AP 3, 15, 27, 39, … is 132 more than its 54th term?
    A. d = 12; 54th term = 3 + 53·12 = 639. Required term = 771. n = (771 − 3)/12 + 1 = 65th.
  3. Q. Find three numbers in AP whose sum is 24 and product is 440.
    A. Take a − d, a, a + d ⇒ 3a = 24 ⇒ a = 8. Product: 8(64 − d²) = 440 ⇒ d² = 9 ⇒ d = ±3. Numbers: 5, 8, 11.