The following recurrence relation can generate a sequence of numbers.
`T_0 = 10, qquad T_(n + 1) = T_n + 3`
The number 13 appears in this sequence as
- `T_1`
- `T_2`
- `T_3`
- `T_10`
- `T_13`
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The following recurrence relation can generate a sequence of numbers.
`T_0 = 10, qquad T_(n + 1) = T_n + 3`
The number 13 appears in this sequence as
`A`
`text(If)\ \ n = 0:`
`T_1` | `= T_0 + 3` |
`= 10 + 3` | |
`= 13` |
`=> A`
A sequence of numbers is generated by the recurrence relation shown below.
`P_0 = 2,quadqquadP_(n + 1) = 3P_n - 1`
What is the value of `P_3`?
`D`
`P_1 = 3P_0 – 1 = 3 xx 2 – 1 = 5`
`P_2 = 3 xx 5 – 1 = 14`
`P_3 = 3 xx 14 – 1 = 41`
`=>\ D`
Consider the recurrence relation shown below.
`A_0 = 3, qquad A_(n + 1) = 2A_n + 4`
The value of `A_3` in the sequence generated by this recurrence relation is given by
`D`
`A_1 = 2A_0 + 4 = 2 xx 3 + 4 = 10`
`A_2 = 2 xx 10 + 4 = 24`
`A_3 = 2 xx 24 + 4`
`=> D`
Consider the recurrence relation below.
`A_0 = 2,\ \ \ \ \ A_(n + 1) = 3 A_n + 1`
The first four terms of this recurrence relation are
`E`
`A_0` | `= 2\ \ (text(given))` |
`A_1` | `= 3(2) + 1 = 7` |
`A_2` | `= 3(7) + 1 = 22` |
`A_3` | `= 3(22) + 1 = 67` |
`=> E`
Paul has to replace 3000 m of fencing on his farm.
Let `F_n` be the length, in metres, of fencing left to replace after `n` weeks.
The difference equation
`F_(n + 1) = 0.95F_n + a\ \ \ \ \ \ F_0 = 3000`
can be used to calculate the length of fencing left to replace after `n` weeks.
In this equation, `a` is a constant.
After one week, Paul still has 2540 m of fencing left to replace.
After three weeks, the length of fencing, in metres, left to replace will be closest to
A. 1310
B. 1380
C. 1620
D. 1690
E. 2100
`D`
`F_(n + 1) = 0.95F_n + a`
`F_1` | `= 0.95F_0 + a` |
`2540` | `= 0.95 xx 3000 + a` |
`:.a` | `= – 310` |
`F_2` | `= 0.95 xx 2540 – 310` |
`= 2103` |
`F_3` | `= 0.95 xx 2103 – 310` |
`= 1687.85` |
`=> D`
Miki is competing as a runner in a half-marathon.
After 30 minutes, his progress in the race is modelled by the difference equation
`K_(n + 1) = 0.99K_n + 250,\ \ \ \ \ \ K_30 = 7550`
where `n ≥ 30` and `K_n` is the total distance Miki has run, in metres, after `n` minutes.
Using this difference equation, the total distance, in metres, that Miki is expected to have run 32 minutes after the start of the race is closest to
A. 7650
B. 7725
C. 7800
D. 7900
E. 8050
`D`
`K_(n + 1) = 0.99K_n + 250`
`text(S)text(ince)\ \ K_30 = 7550,`
`K_31` | `= 0.99 xx 7550 + 250` |
`= 7724.5` | |
`K_32` | `= 0.99 xx 7724.5 + 250` |
`= 7897.255` |
`=> D`
Paula started a stamp collection. She decided to buy a number of new stamps every week.
The number of stamps bought in the `n`th week, `t_n`, is defined by the difference equation
`t_n = t_(n-1) + t_(n-2)\ \ \ text(where)\ \ \ t_1 = 1 and t_2 = 2`
The total number of stamps in her collection after five weeks is
A. `8`
B. `12`
C. `15`
D. `19`
E. `24`
`D`
`t_1=1,\ \ \ t_2=2\ \ \ text{(given)}`
`:. t_3` | `=t_2 + t_1 = 2+1=3` |
`t_4` | `=3+2=5` |
`t_5` | `=5+3= 8` |
`:.\ text(Total stamps after 5 weeks)`
`=1+2+3+5+8=19`
`rArr D`
A difference equation is defined by
`f_(n+1) - f_n = 5\ \ \ \ \ text (where)\ \ f_1 =– 1`
The sequence `f_1, \ f_2, \ f_3, ...` is
A. `5, 4, 3\ …`
B. `4, 9, 14\ …`
C. `– 1, – 6, – 11\ …`
D. `– 1, 4, 9\ …`
E. `– 1, 6, 11\ …`
`D`
`text(Using)\ \ \ f_(n+1) = f_n +5`
`text(Consider)\ \ f_1 = – 1,`
`f_2= – 1 + 5=4`
`f_3 = 4+5=9`
`rArr D`
The following information relates to Parts 1 and 2.
A farmer plans to breed sheep to sell.
In the first year she starts with 50 breeding sheep.
During the first year, the sheep numbers increase by 84%.
At the end of the first year, the farmer sells 40 sheep.
Part 1
How many sheep does she have at the start of the second year?
A. 2
B. 42
C. 52
D. 84
E. 92
Part 2
If `S_n` is the number of sheep at the start of year `n`, a difference equation that can be used to model the growth in sheep numbers over time is
A. `S_(n+1) = 1.84S_n - 40` | `\ \ \ \ \ text(where)\ \ S_1 = 50` | |
B. `S_(n+1) = 0.84S_n - 50` | `\ \ \ \ \ text(where)\ \ S_1 = 40` | |
C. `S_(n+1) = 0.84S_n - 40` | `\ \ \ \ \ text(where)\ \ S_1 = 50` | |
D. `S_(n+1) = 0.16S_n - 50` | `\ \ \ \ \ text(where)\ \ S_1 = 40` | |
E. `S_(n+1) = 0.16S_n - 40` | `\ \ \ \ \ text(where)\ \ S_1 = 50` |
`text (Part 1:)\ C`
`text (Part 2:)\ A`
`text (Part 1)`
`text(Number at the start of the 2nd year)`
`=50 + (84text(%) xx 50) – 40`
`=52`
`rArr C`
`text (Part 2)`
`text(If sheep numbers increase by 84% from the start of)`
`text(each year, and 40 sheep are then sold,)`
`S_(n+1)` | `=S_n + 0.84S_n – 40` |
`=1.84 S_n – 40` |
`rArr A`
The initial rate of pay for a job is $10 per hour.
A worker’s skill increases the longer she works on this job. As a result, the hourly rate of pay increases each month.
The hourly rate of pay in the `n`th month of working on this job is given by the difference equation
`S_(n+1) = 0.2 xx S_n+15\ \ \ \ \ \ S_1 = 10`
The maximum hourly rate of pay that the worker can earn in this job is closest to
A. $3.00
B. $12.00
C. $12.50
D. $18.75
E. $75.00
`D`
`text(Maximum rate)\ S_(max)\ text(occurs when)`
`S_(n+1)` | `=S_n` |
`S_(max)` | `=0.2×S_(max)+15` |
`0.8 xx S_(max)` | `=15` |
`S_(max)` | `=15/0.8` |
`=18.75` |
`=> D`
The difference equation `u_(n + 1) = 4u_n - 2` generates a sequence.
If `u_2 = 2`, then `u_4` will be equal to
A. 4
B. 8
C. 22
D. 40
E. 42
`C`
`u_(n+1)` | `= 4u_n – 2` | |
`∴ u_3` | `= 4u_2 – 2` | |
`= 4 xx 2 – 2\ \ text{(given}\ u_2 = 2 text{)}` | ||
`= 6` | ||
`∴ u_4` | `= 4u_3 – 2` | |
`= 4 xx 6 – 2` | ||
`= 22` |
`=> C`
The difference equation
`t_(n+2) = t_(n+1) + t_n` where `t_1 = a` and `t_2 = 7`
generates a sequence with `t_5 = 27`.
The value of `a` is
A. 0
B. 1
C. 2
D. 3
E. 4
`D`
`t_(n+2) = t_(n+1) + t_n\ \ text(where)\ \ t_1 = a\ \ text(and)\ \ t_2 = 7`
`text(Calculating this equation from)\ \ n = 1,`
`t_3 ` | ` = t_2 + t_1` |
` = 7 + a` | |
`t_4 ` | ` = t_3 + t_2` |
` = 7 + a + 7` | |
` = 14 + a` | |
`t_5` | ` = t_4 + t_3` |
`:. 27 ` | ` = 14 + a + 7 + a` |
`2a ` | ` = 6` |
`a ` | ` = 3` |
`=> D`
The first term of a Fibonacci-related sequence is `p`.
The second term of the same Fibonacci-related sequence is `q`.
The difference in value between the fourth and fifth terms of this sequence is
A. `p - q`
B. `q - p`
C. `p + q`
D. `p + 2q`
E. `2p + 3q`
`C`
`text(Fibonacci sequence general form is)`
`t_(n+2) = t_(n+1) + t_n`
`t_1 = p`
`t_2 = q`
`t_3 = p + q`
`t_4 = (p + q) + q = p + 2q`
`t_5 = p + 2q + (p + q) = 2p + 3q`
`∴ t_5 – t_4` | `= 2p + 3q – (p + 2q)` |
`= p + q` |
`=> C`
Three years after observations began, 12 300 birds were living in a wetland.
The number of birds living in the wetland changes from year to year according to the difference equation
`t_(n+ 1) = 1.5t_n - 3000, quad quad t_3 = text (12 300)`
where `t_n` is the number of birds observed in the wetland `n` years after observations began.
The number of birds living in the wetland one year after observations began was closest to
A. `8800`
B. `9300`
C. `10\ 200`
D. `12\ 300`
E. `120\ 175`
`A`
`t_(n+1)` | `= 1.5t_n – 3000, and t_3 =12\ 300` |
`:. t_3` | `= 1.5t_2 – 3000` |
`12\ 300` | `= 1.5t_2 – 3000` |
`1.5t_2` | `= 15\ 300` |
`t_2` | `= 10\ 200` |
`text(Similarly,)`
`t_2` | `= 1.5t_1 – 3000` |
`10\ 200` | `= 1.5t_1 – 3000` |
`1.5t_1` | `= 13\ 200` |
`:. t_1` | `= 8800` |
`rArr A`
1, 9, 10, 19, 29, . . .
The sixth term of the Fibonacci-related sequence shown above is
A. 30
B. 39
C. 40
D. 48
E. 49
`D`
`text (Sequence is 1, 9, 10, 19, 29…)`
`text (6th term)` | `= 4text (th + 5th)` |
`=19 + 29` | |
`= 48` |
`rArr D`
The first six terms of a Fibonacci related sequence are shown below.
4, 7, 11, 18, 29, 47, ...
The next term in the sequence is
A. 58
B. 65
C. 76
D. 94
E. 123
`C`
`text(Fibonacci general term,)`
`T_(n+2) = T_(n+1) + T_n`
`:.\ text(Next term)` | `= 29 + 47` |
`= 76` |
`rArr C`
A sequence is generated by the difference equation
`t_(n+1)=2 xx t_n,\ \ \ \ \ t_1=1`
The `n`th term of this sequence is
A. `t_n=1×2^(n-1)`
B. `t_n=1+2^(n-1)`
C. `t_n=1+2×(n-1)`
D. `t_n=2+(n-1)`
E. `t_n=2+1^(n-1)`
`A`
`t_2` | `=2 xx t_1 = 2` |
`t_3` | `=2 xx t_2 = 2^2` |
`t_4` | `=2 xx t_3 = 2^3` |
`t_5` | `=2 xx t_4 = 2^4` |
`vdots`
`t_n= 1 xx 2^(n-1)`
`=> A`