D: Sequences and series: Understand and work with geometric sequences and series including the formulae for the n th term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of r < 1 ; modulus notation.

The idea

A geometric sequence has a constant ratio r between consecutive terms. The nth term is ar^(n-1), where a is the first term. The sum of the first n terms is a(1-r^n)/(1-r) or a(r^n-1)/(r-1) for r≠1. A geometric series converges when |r|<1, and then the sum to infinity is a/(1-r). Modulus notation |r|<1 means -1<r<1; the inequality alone permits r=0, but a geometric common ratio is taken to be nonzero, so r≠0 is an additional convention rather than a consequence of the modulus. To work with these, identify a and r, choose the correct formula, and substitute. For example, in 3, 6, 12, 24, ... a=3 and r=2, so the 10th term is 3×2^9=1536 and the sum of the first 10 terms is 3(2^10-1)/(2-1)=3069. For a convergent series such as 8, 4, 2, 1, ..., a=8 and r=0.5, so S∞=8/(1-0.5)=16. Recognise convergence and use modulus notation correctly.

What earns the marks

• Correctly identifies the first term a and common ratio r from a geometric sequence or from contextual information, including when r is negative or fractional.
• Uses the nth term formula ar^(n-1) accurately to find a specified term or to form an equation when a term is known.
• Applies the finite sum formula S_n = a(1-r^n)/(1-r) or S_n = a(r^n-1)/(r-1) correctly, choosing the version that avoids unnecessary negative signs.
• Determines whether a geometric series converges by checking |r|<1, and calculates the sum to infinity using S∞ = a/(1-r) when convergence is established.
• Interprets modulus notation |r|<1 correctly as -1<r<1, distinguishing this from the weaker condition r<1, and uses it to justify convergence or to find possible values of r.

Common mistakes and how to put them right

• Using r^n instead of r^(n-1) in the nth term formula.

  Put it right by remembering that the first term corresponds to n=1, so the exponent is n-1.

• Applying the sum to infinity formula when |r|≥1.

  Put it right: First checking the modulus condition and stating that the series does not converge if |r|≥1.

• Misinterpreting |r|<1 as r<1 only.

  Put it right by treating it as -1<r<1, which includes negative ratios with magnitude less than 1, and note that the modulus inequality alone does not exclude r=0.

Exam tips

• Write down a and r before substituting into any formula, and state the condition |r|<1 explicitly when using the sum to infinity.
• When a question involves a geometric series in context, check whether the ratio is constant and whether the series is finite or infinite before choosing a formula.
• Use the modulus notation correctly in written answers; for example, write |r|<1 rather than just r<1 when justifying convergence.