Arithmetic sequences and series

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What this lesson covers

  • The nth term is a + (n − 1)d.
  • The sum of n terms is (n/2)(2a + (n − 1)d).
  • Two known terms are enough to find a and d.
Key idea

The two formulas

  • a is the first term and d is the common difference.
  • tₙ = a + (n − 1)d.
  • Sₙ = (n/2)(2a + (n − 1)d).
Worked example

5, 9, 13, ...

  • Find the 20th term and the sum of 20 terms.
  1. a = 5, d = 4
  2. t₂₀ = 5 + 19 × 4 = 81
  3. S₂₀ = 10(10 + 19 × 4)
  4. S₂₀ = 860
Worked example

Given t₅ = 14 and t₁₂ = 42

  • The gap between the terms holds 7 lots of d.
  1. 7d = 28, so d = 4
  2. a + 4(4) = 14
  3. a = −2

Common mistakes

  • Using n × d in the nth term.
    The first term has had no differences added, so it is (n − 1)d.
  • Mixing up the term and the sum.
    tₙ is one term and Sₙ adds them all.

Your turn

  • 1. For 3, 10, 17, ... find t₁₅.
  • 2. Find S₁₅.
Show answer

1. 3 + 14 × 7 = 101 2. 7.5(6 + 98) = 780

Summary

What to take away

  • Same difference means arithmetic.
  • Use (n − 1), never n.

More in AS 91258

Arithmetic sequences and seriesNotesGeometric sequences and seriesNotesSigma notationNotesGrowth and financial problemsNotes

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