AQA Paper 3 2019 June — Question 3 1 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSimple recurrence evaluation
DifficultyEasy -1.2 This is a 1-mark multiple choice question requiring students to identify an increasing sequence by testing each formula with u₁=1. The first option gives u₂=2, u₃=1.5 (decreasing), the second gives u₂≈1.1, u₃≈1.19 (increasing), making it straightforward pattern recognition with minimal calculation. Below average difficulty as it requires only basic substitution and comparison, not proof or deep understanding.
Spec1.04e Sequences: nth term and recurrence relations1.04f Sequence types: increasing, decreasing, periodic

Given \(u_1 = 1\), determine which one of the formulae below defines an increasing sequence for \(n \geq 1\) Circle your answer. [1 mark] \(u_{n+1} = 1 + \frac{1}{u_n}\) \quad \(u_n = 2 - 0.9^{n-1}\) \quad \(u_{n+1} = -1 + 0.5u_n\) \quad \(u_n = 0.9^{n-1}\)

Question 3:
AnswerMarks Guidance
3Circles the correct response 2.2a
n
u =2−0.9n−1
n
AnswerMarks Guidance
Total1
QMarking instructions AO
Question 3:
3 | Circles the correct response | 2.2a | B1 | u =2−0.9n−1
n
u =2−0.9n−1
n
Total | 1
Q | Marking instructions | AO | Mark | Typical solution
Given $u_1 = 1$, determine which one of the formulae below defines an increasing sequence for $n \geq 1$

Circle your answer.
[1 mark]

$u_{n+1} = 1 + \frac{1}{u_n}$ \quad $u_n = 2 - 0.9^{n-1}$ \quad $u_{n+1} = -1 + 0.5u_n$ \quad $u_n = 0.9^{n-1}$

\hfill \mbox{\textit{AQA Paper 3 2019 Q3 [1]}}