Alternation around a fixed point · source reference

Printed question

(a) A sequence of real numbers $u_1,u_2,u_3,\ldots$ is such that

$$u_{n+1}=3-0.5u_n,\quad n\ge1.$$

It is given that $u_1=c$ where $c$ is a constant.

(i) Describe how the sequence behaves when (A) $c=5$ [1], (B) $c=2$. [1]

(ii) Find the value of $c$ for which $2u_3=-5u_2$. [3]

(b) Another sequence is defined by $v_1=p,v_2=2$, where $p$ is a constant, and

$$v_{n+2}=2v_n+v_{n+1}-1,\quad n\ge1.$$

(i) If $v_3+1=2u_2-u_1$, find a relationship between $c$ and $p$. [2]

(ii) Find the value of $p$ for which $v_5=77$. [2]

Printed working

(a)(i)(A) $c=5$: the sequence initially decreases and subsequently alternates, and converges to 2.

(B) $c=2$: it is a constant with a value of 2.

(a)(ii)

$$\begin{aligned}u_1&=c,\\u_2&=3-0.5u_1=3-0.5c,\\u_3&=3-0.5u_2=3-0.5(3-0.5c)=1.5+0.25c,\\2(1.5+0.25c)&=-5(3-0.5c),\\3+0.5c&=-15+2.5c,\\2c&=18,\quad c=9.\end{aligned}$$

(b)(i)

$$v_3+1=2u_2-u_1\Rightarrow2v_1+v_2-1+1=2u_2-u_1,$$ $$2p+2=2(3-0.5c)-c=6-2c\Rightarrow c+p=2.$$

(b)(ii)

$$\begin{aligned}v_3&=2v_1+v_2-1=2p+1,\\v_4&=2v_2+v_3-1=2(2)+(2p+1)-1=2p+4,\\v_5&=2v_3+v_4-1=2(2p+1)+(2p+4)-1=6p+5.\end{aligned}$$

$6p+5=77\Rightarrow p=12$.

Teacher notes

Using GC:

1. Press mode, scroll to SEQ (sequence mode).
2. Press Y=. You will see nMin = 1. Enter $u(n)=3-0.5u(n-1)$. Press 2nd 7 for “u” and X,T,θ,n for “n”. For $c=5$, put $u(1)=5,u(2)=0.5$. For $c=2$, put $u(1)=2,u(2)=2$.
3. Press 2nd graph to obtain a table of sequence values.

Source tables and calculator displays

Calculator table for $c=5$ (display precision retained):

$n$	$u$
10	1.9941
11	2.0029
12	1.9985
13	2.0007
14	1.9996
15	2.0002
16	1.9999
17	2
18	2
19	2
20	2

For $c=2$, the second calculator table lists $n=1,2,\ldots,11$, all with $u=2$. The displayed final 2s in the first table are rounded; they do not mean the sequence becomes exactly constant.

Printed answer key

(a)(i)(A) Initially decreases and subsequently alternates, converging to 2. (B) Constant value 2. (a)(ii) 9; (b)(i) $c+p=2$; (ii) 12

Mathematical review and qualifications

(a)(i) $u_{n+1}-2=-\frac12(u_n-2)$, hence $u_n=2+(c-2)(-1/2)^{n-1}$. For $c=5$, the terms $5,0.5,2.75,1.625,\ldots$ alternate around 2 and converge to 2; for $c=2$ every term is 2.

(ii) $u_2=3-c/2$, $u_3=3/2+c/4$. Thus $3+c/2=-15+5c/2$ gives $c=9$.

(b) $v_3=2p+1$, $v_4=2p+4$, $v_5=6p+5$. (i) $2p+2=6-2c$, so $c+p=2$. (ii) $6p+5=77$ gives $p=12$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2023/DHS/Prelim/II/2] Question PDF pages 1; solution PDF pages 2, 3. The original files remain private QA records outside this native content.