Understand and work with arithmetic sequences and series, including the…

The idea

An arithmetic sequence adds a constant common difference d to each term, so a, a + d, a + 2d, ... The nth term is uₙ = a + (n − 1)d, where a is the first term. The sum of the first n terms is Sₙ = n/2[2a + (n − 1)d], equivalently Sₙ = n/2(a + l) when the last term l is known. For example, 5, 8, 11, 14, ... has a = 5 and d = 3, so u₁₀ = 5 + 9 × 3 = 32 and S₁₀ = 10/2(2 × 5 + 9 × 3) = 5 × 37 = 185. To work with these, identify a and d, substitute into the correct formula, and solve the resulting linear or quadratic equation. Recognise that d may be negative or a fraction, and that n must be a positive integer.

What earns the marks

• States uₙ = a + (n − 1)d and substitutes the correct a, d and n to find a term or to form an equation.
• Uses Sₙ = n/2[2a + (n − 1)d] or Sₙ = n/2(a + l), choosing the form that matches the given information.
• Finds a and d from contextual or algebraic information, for example from two given terms or from simultaneous equations.
• Solves equations arising from uₙ or Sₙ, including quadratic equations in n, and rejects non-integer or non-positive values of n.
• Interprets the result in context, such as identifying which term number gives a stated value or the total after a given number of terms.

Common mistakes and how to put them right

• Using uₙ = a + nd instead of a + (n − 1)d.

  Put it right by subtracting 1 from n before multiplying by d.

• Treating d as always positive.

  Put it right by calculating d = u₂ − u₁ and retaining its sign.

• Accepting a negative or fractional value of n from a quadratic.

  Put it right by rejecting it because n counts terms and must be a positive integer.

Exam tips

• Write down a and d explicitly before substituting, so arithmetic slips are easier to spot.
• When two terms are given, form two equations and solve simultaneously rather than guessing a and d.
• Check a sum by adding the first few terms manually or by using Sₙ = n/2(a + l) as a second method.