Shift an affine recurrence into a GP · source reference

Printed question

The $n$th term of a sequence is denoted by $a_n$. The sequence is defined by

$$a_{n+1}=pa_n+q\quad(n\ge1),$$

where $p,q$ are constants. The first three terms are $a_1=0,a_2=2,a_3=7$.

(i) Find $p,q$. [2]

(ii) Express $a_n$ in terms of $n$, for $n\ge3$. [4]

Printed working

(i) $a_2=pa_1+q\Rightarrow q=2$.

$a_3=pa_2+q\Rightarrow7=2p+2\Rightarrow p=\frac52$.

(ii) Given $a_{n+1}=\frac52a_n+2$,

$$\begin{aligned}a_2&=2,\\a_3&=\frac52a_2+2=\frac52(2)+2,\\a_4&=\frac52\left[\frac52(2)+2\right]+2=\left(\frac52\right)^2(2)+\frac52(2)+2,\\a_n&=\left(\frac52\right)^{n-2}(2)+\left(\frac52\right)^{n-3}(2)+\cdots+\frac52(2)+2\\&=2\left[1+\frac52+\left(\frac52\right)^2+\cdots+\left(\frac52\right)^{n-2}\right]\\&=\frac{2[(5/2)^{n-1}-1]}{5/2-1}=\frac43\left[\left(\frac52\right)^{n-1}-1\right],\quad n\ge3.\end{aligned}$$

Teacher notes

Observe the pattern in the series: do not simplify to numerical values. Leave $\left(\frac52\right)^2(2)+\frac52(2)+2$ instead of $\frac{25}2+5+2=\frac{39}2$.

The bracketed series is a GP with first term 1 and common ratio $\frac52$.

Printed answer key

(i) $p=5/2,q=2$; (ii) $a_n=\frac43[(5/2)^{n-1}-1]$, $n\ge3$

Mathematical review and qualifications

(i) $2=p(0)+q$ gives $q=2$; $7=2p+2$ gives $p=5/2$.

(ii) Set $b_n=a_n+4/3$. Then $b_{n+1}=(5/2)b_n$, $b_1=4/3$. Therefore $a_n=\frac43[(5/2)^{n-1}-1]$, valid in particular for $n\ge3$. Substitution gives $0,2,7$ as required.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/TJC/Prelim/II/1)] Question PDF pages 1; solution PDF pages 3, 4. The original files remain private QA records outside this native content.