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`?
- 2
- 5
- 11
- 41
- 122
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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`
The first five terms of a sequence are 2, 6, 22, 86, 342 …
The recurrence relation that generates this sequence could be
`D`
`text(Using trial and error,)`
`text(Consider option)\ D:`
`P_1 = 4 xx 2 – 2 = 6`
`P_2 = 4 xx 6 – 2 = 22`
`P_3 = 4 xx 22 – 2 = 86\ \ text(etc…)`
`=> 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`
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 difference equation
`t_(n+1) = at_n + 6 quad text (where) quad t_1 = 5`
generates the sequence
`5, 21, 69, 213\ …`
The value of `a` is
A. – 1
B. 3
C. 4
D. 15
E. 16
`B`
`t_2` | `=a t_1 +6` |
`:. 21` | `=5a + 6` |
`a` | `=3` |
`rArr B`
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`
A patient takes 15 milligrams of a prescribed drug at the start of each day.
Over the next 24 hours, 85% of the drug in his body is used. The remaining 15% stays in his body.
Let `D_n` be the number of milligrams of the drug in the patient’s body immediately after taking the drug at the start of the `n`th day.
A difference equation for determining `D_(n+1)`, the number of milligrams in the patient’s body immediately after taking the drug at the start of the `n+1`th day, is given by
A. `D_(n + 1) = 85 D_n + 15` | `D_1 = 15` |
B. `D_(n + 1) = 0.85 D_n + 15` | `D_1 = 15` |
C. `D_(n + 1)= 0.15 D_n + 15` | `D_1 = 15` |
D. `D_(n + 1)= 0.15 D_n + 0.85` | `D_1 = 15` |
E. `D_(n + 1)= 15 D_n + 85` | `D_1 = 15` |
`C`
`D_1=15`
`text(85% of the drug is used up before the second dose.)`
`D_2` | `=0.15 D_1 + 15\ \ \ text{(drug left from 1st day + new dose)}` |
`D_3` | `= 0.15 D_2 + 15\ \ \ text{(drug left from 2nd day + new dose)}` |
`vdots` | |
`D_(n+1)` | `=0.15 D_n+15` |
`=> C`
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`
Let `P_2011` be the number of pairs of shoes that Sienna owns at the end of 2011.
At the beginning of 2012, Sienna plans to throw out the oldest 10% of pairs of shoes that she owned in 2011.
During 2012 she plans to buy 15 new pairs of shoes to add to her collection.
Let `P_2012` be the number of pairs of shoes that Sienna owns at the end of 2012.
A rule that enables `P_2012` to be determined from `P_2011` is
A. `P_2012 = 1.1 P_2011 + 15`
B. `P_2012 = 1.1 (P_2011 + 15)`
C. `P_2012 = 0.1 P_2011 + 15`
D. `P_2012 = 0.9 (P_2011 + 15)`
E. `P_2012 = 0.9 P_2011 + 15`
`E`
`text(By throwing out 10%, Sienna keeps 90% of her)`
`text{her 2011 shoes (or 0.9} \ P_2011 text{) and then adds 15.}`
`:. P_(2012) = 0.9\ P_2011 + 15`
`=> E`
Sam takes a tablet containing 200 mg of medicine once every 24 hours.
Every 24 hours, 40% of the medicine leaves her body. The remaining 60% of the medicine stays in her body.
Let `D_n` be the number of milligrams of the medicine in Sam’s body immediately after she takes the `n`th tablet.
The difference equation that can be used to determine the number of milligrams of the medicine in Sam’s body immediately after she takes each tablet is shown below.
`D_(n + 1) = 0.60D_n + 200,\ \ \ \ \ \ D_1 = 200`
Which one of the following statements is not true?
A. The number of milligrams of the medicine in Sam’s body never exceeds 500.
B. Immediately after taking the third tablet, 392 mg of the medicine is in Sam’s body.
C. The number of milligrams of the medicine that leaves Sam’s body during any 24-hour period will always be less than 200.
D. The number of milligrams of the medicine that leaves Sam’s body during any 24-hour period is constant.
E. If Sam stopped taking the medicine after the fifth tablet, the amount of the medicine in her body would drop to below 200 mg after a further 48 hours.
`D`
`text(Consider A:)`
`text(Maximum medicine in body when)`
`D_(n+1)` | `= D_n` |
`x` | `= 0.6x + 200` |
`0.4x` | `= 200` |
`x` | `=500,\ =>\ text(A true)` |
`text(Consider B:)`
`D_1 = 200`
`D_2 = 0.6(200) + 200 = 320`
`D_3 = 0.6(320) + 200 = 392,\ =>\ text(B true)`
`text(Consider C:)`
`text{Max medicine never exceeds 500 mg (from A),}\ =>\ text(C true)`
`text(Consider D:)`
`text(Medicine leaving body is 40% of a changing number,)\ =>\ text(D not true)`
`text(Consider E:)`
`D_4 = 0.6(392) + 200 = 435.2`
`D_5 = 0.6(435.2) + 200 = 461.12`
`D_6 = 0.6(461.12) = 276.67`
`D_7 = 0.6(276.672) = 166.00,\ =>\ text(E true)`
`=>D`
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`