A sequence of numbers is generated by the recurrence relation shown below.

\(T_0=5, \quad T_{n+1}=-T_n\)

The value of \(T_2\) is

- \(-10\)
- \(-5\)
- \(0\)
- \(5\)
- \(10\)

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A sequence of numbers is generated by the recurrence relation shown below.

\(T_0=5, \quad T_{n+1}=-T_n\)

The value of \(T_2\) is

- \(-10\)
- \(-5\)
- \(0\)
- \(5\)
- \(10\)

Show Answers Only

\(D\)

Show Worked Solution

\(T_1 = -T_0 = -5\)

\(T_2 = -T_1 = 5\)

\(\Rightarrow D\)

`A_n` is the `n`th term in a sequence.

Which one of the following expressions does not define a geometric sequence?

A. `A_(n + 1) = n` |
`\ \ \ \ A_0 = 1` |

B. `A_(n + 1) = 4` |
`\ \ \ \ A_0 = 4` |

C. `A_(n + 1) = A_n + A_n` |
`\ \ \ \ A_0 = 3` |

D. `A_(n + 1) = –A_n` |
`\ \ \ \ A_0 = 5` |

E. `A_(n + 1) = 4A_n` |
`\ \ \ \ A_0 = 2` |

Show Answers Only

`A`

Show Worked Solution

`text(Test all options by looking at the first)`

`text(3 terms that each produces.)`

`text(Consider)\ A,`

`A_0=1, \ A_1=0, \ A_2=1`

`text(There is no common ratio in this sequence.)`

`text(All other options can be shown to have a common ratio.)`

`=> A`

A town has a population of 200 people when a company opens a large mine.

Due to the opening of the mine, the town’s population is expected to increase by 50% each year.

Let `P_n` be the population of the town `n` years after the mine opened.

The expected growth in the town’s population can be modelled by

A. `P_(n + 1) = P_n + 100` |
`\ \ \ \ \ P_0 = 200` |

B. `P_(n + 1) = P_n + 100` |
`\ \ \ \ \ P_1= 300` |

C. `P_(n + 1) = 0.5P_n` |
`\ \ \ \ \ P_0 = 200` |

D. `P_(n + 1) = 1.5P_n` |
`\ \ \ \ \ P_0 = 300` |

E. `P_(n + 1) = 1.5P_n` |
`\ \ \ \ \ P_1 = 300` |

Show Answers Only

`E`

Show Worked Solution

`text(After 1 year,)`

`P_1` | `= 1.5 xx P_0` |

`= 1.5 xx 200` | |

`=300` |

`=> E`

*The following information relates to Parts 1 and 2.*

The number of waterfowl living in a wetlands area has decreased by 4% each year since 2003.

At the start of 2003 the number of waterfowl was 680.

**Part 1**

If this percentage decrease continues at the same rate, the number of waterfowl in the wetlands area at the start of 2008 will be closest to

**A.** 532

**B.** 544

**C.** 554

**D.** 571

**E.** 578

**Part 2**

`W_n`* *is the number of waterfowl at the start of the* *`n`th year.

Let `W_1 = 680.`

The rule for a difference equation that can be used to model the number of waterfowl in the wetlands area over time is

**A.** `W_(n+1) = W_n - 0.04n`

**B. ** `W_(n+1) = 1.04 W_n`

**C.** `W_(n+1) = 0.04 W_n`

**D. ** `W_(n+1) = -0.04 W_n`

**E.** `W_(n+1) = 0.96 W_n`

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ E`

Show Worked Solution

`text (Part 1)`

`text(After 1 year, number of waterfowls)`

`=680 – 4/100 xx 680`

`=680\ (0.96)^1`

`text(After 2 years)\ = 680\ (0.96)^2`

`vdots`

`text{After 5 years (in 2008)}`

`=680\ (0.96)^5 =554.45…`

`rArr C`

`text (Part 2)`

`text(Sequence is geometric where)\ \ r=0.96`

`:. W_(n+1)/W_n` | `=0.96` |

`W_(n+1)` | `=0.96 W_n` |

`rArr E`

In 2008, there are 800 bats living in a park.

After 2008, the number of bats living in the park is expected to increase by 15% per year.

Let `Β_n` represent the number of bats living in the park `n` years after 2008.

A difference equation that can be used to determine the number of bats living in the park `n` years after 2008 is

A. `B_n=1.15B_(n-1)-800` |
`\ \ \ \ \ B_0=2008` |

B. `B_n=B_(n-1)+1.15xx800` |
`\ \ \ \ \ B_0=2008` |

C. `B_n=B_(n-1)-0.15xx800` |
`\ \ \ \ \ B_0=800` |

D. `B_n=0.15B_(n-1)` |
`\ \ \ \ \ B_0=800` |

E. `B_n=1.15B_(n-1)` |
`\ \ \ \ \ B_0=800` |

Show Answers Only

`E`

Show Worked Solution

`B_0=800`

`B_1= B_0 + 15 text(%) xx B_0=1.15 B_0`

`B_2= 1.15B_1`

`vdots`

`B_n=1.15 B_(n-1)`

`=> E`

Consider the following sequence.

`2,\ 1,\ 0.5\ …`

Which of the following difference equations could generate this sequence?

A. |
`t_(n + 1) = t_n - 1` | `t_1 = 2` |

B. |
`t_(n + 1) = 3 - t_n` | `t_1 = 2` |

C. |
`t_(n + 1) = 2 × 0.5^(n – 1)` | `t_1 = 2` |

D. |
`t_(n + 1) = - 0.5t_n + 2` | `t_1 = 2` |

E. |
`t_(n + 1) = 0.5t_n` | `t_1 = 2` |

Show Answers Only

`E`

Show Worked Solution

`text(Sequence is)\ \ 2, 1, 0.5, …`

`=>\ text(Geometric sequence where common ratio = 0.5)`

`∴\ text(Difference equation is)`

`t_(n + 1) = 0.5t_n`

`=> E`

A poultry farmer aims to increase the weight of a turkey by 10% each month.

The turkey’s weight, `T_n`, in kilograms, after `n` months, would be modelled by the rule

**A.** `T_(n + 1) = T_n + 10`

**B. **`T_(n + 1) = 1.1T_n + 10`

**C.** `T_(n + 1) = 0.10T_n`

**D.** `T_(n + 1) = 10T_n`

**E.** `T_(n + 1) = 1.1T_n`

Show Answers Only

`E`

Show Worked Solution

`T_2` | `=1.1T_1` |

`T_3` | `= 1.1T_2` |

`vdots` | |

`T_(n+1)` | `= 1.1T_n` |

`rArr E`

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)`

Show Answers Only

`A`

Show Worked Solution

`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`