Here is a rule that can be used to build a sequence of numbers once a starting number is chosen: Each number is two times three less than the previous number.

a. Starting with the number 0, build a sequence of 5 numbers.
b. Starting with the number 3, build a sequence of 5 numbers.
c. Can you choose a starting point so that the first 5 numbers in your sequence are all positive? Explain your reasoning.

Here the sequence is governed by the rule:
[tex]\sf a_{n} = 2(a_{n-1} - 3)[/tex]
[tex]\textsf {where $a_n$ is the nth term and $a_{n-1}$ is the previous term}}[/tex]

1) Starting with a₁ = 0 we get
a₂ = 2(a₁ - 3) = 2(0 - 3) = -6
a₃ = 2(a₂ - 3) = 2(-6 -3) = 2(-9) = -18
a₄ = 2(a3 - 3) = 2(-18 - 3) = 2(-21) = -42
a₅ = 2(a₄ - 3) = 2(-42- 3) = 2(-45) = -90

So the first 5 numbers in the sequence starting with 0 are:
0, -6, -18, -21, -90

2) I will approach this slightly differently

We have the relationship:
[tex]\sf a_{n} = 2(a_{n-1} - 3)[/tex]

Expanding the brackets we get
[tex]\sf a_{n} = 2a_{n-1} - 6[/tex]

We have a₁ = 3
a₂ = 2(a₁) - 6 = 2(3) - 6 = 6 - 6 = 0
a₃ = 2(a₂) - 6 = 2(0) - 6 = -6
a₄ = 2(a₃) - 6 = 2(-6) - 6 = -18
a₅ = 2(a₄ ) - 6 = 2(-18) - 6 = -42

The sequence starting with 3 is:
3, 0, -6, -18, -42

Both methods work exactly the same, some students find expanding the brackets easier to deal with

c. We can see that as a₁ increases, the number of negative numbers decreases by 2. So, in order for all numbers to be positive (> 0) we should start with a sufficiently large number so that all numbers are positive

The smallest such number is 6.
When a₁ = 6, the sequence is all 6s because
a₂ = 2(6) - 6 = 12 - 6 = 6

So the next number is a₃ = 12(6) - 6 = 6 and so on

So any starting point ≥ 6 will yield positive values. At 5 the sequence would be: 5, 4, 2, -2, -10 so starting point of 5 will not work