Math worksheet for generating sequences from the nth term, including sections with algebraic expressions and tables for calculating terms.

Let's solve each section of the worksheet "Generating Sequences from the Nth Term" step by step. We'll calculate the required terms for each sequence by substituting values of $ n $ into the given formulas.

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✔ Section A: Find the first five terms (n = 1 to 5)

We substitute $ n = 1, 2, 3, 4, 5 $ into each expression:

| Expression | n=1 | n=2 | n=3 | n=4 | n=5 |
|-----------|-----|-----|-----|-----|-----|
| $ n + 2 $ | $ 1+2 = 3 $ | $ 2+2 = 4 $ | $ 3+2 = 5 $ | $ 4+2 = 6 $ | $ 5+2 = 7 $ |
| $ n - 7 $ | $ 1-7 = -6 $ | $ 2-7 = -5 $ | $ 3-7 = -4 $ | $ 4-7 = -3 $ | $ 5-7 = -2 $ |
| $ 4n $ | $ 4(1) = 4 $ | $ 4(2) = 8 $ | $ 4(3) = 12 $ | $ 4(4) = 16 $ | $ 4(5) = 20 $ |
| $ 3n + 1 $ | $ 3(1)+1 = 4 $ | $ 3(2)+1 = 7 $ | $ 3(3)+1 = 10 $ | $ 3(4)+1 = 13 $ | $ 3(5)+1 = 16 $ |
| $ 5n - 8 $ | $ 5(1)-8 = -3 $ | $ 5(2)-8 = 2 $ | $ 5(3)-8 = 7 $ | $ 5(4)-8 = 12 $ | $ 5(5)-8 = 17 $ |
| $ 2n + 0.5 $ | $ 2(1)+0.5 = 2.5 $ | $ 2(2)+0.5 = 4.5 $ | $ 2(3)+0.5 = 6.5 $ | $ 2(4)+0.5 = 8.5 $ | $ 2(5)+0.5 = 10.5 $ |
| $ 12n - 1.5 $ | $ 12(1)-1.5 = 10.5 $ | $ 12(2)-1.5 = 22.5 $ | $ 12(3)-1.5 = 34.5 $ | $ 12(4)-1.5 = 46.5 $ | $ 12(5)-1.5 = 58.5 $ |

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✔ Section B: Find the first five terms

| Expression | n=1 | n=2 | n=3 | n=4 | n=5 |
|-----------|-----|-----|-----|-----|-----|
| $ 2 - n $ | $ 2-1 = 1 $ | $ 2-2 = 0 $ | $ 2-3 = -1 $ | $ 2-4 = -2 $ | $ 2-5 = -3 $ |
| $ 13 - n $ | $ 13-1 = 12 $ | $ 13-2 = 11 $ | $ 13-3 = 10 $ | $ 13-4 = 9 $ | $ 13-5 = 8 $ |
| $ -3n $ | $ -3(1) = -3 $ | $ -3(2) = -6 $ | $ -3(3) = -9 $ | $ -3(4) = -12 $ | $ -3(5) = -15 $ |
| $ 1 - 2n $ | $ 1-2(1) = -1 $ | $ 1-2(2) = -3 $ | $ 1-2(3) = -5 $ | $ 1-2(4) = -7 $ | $ 1-2(5) = -9 $ |
| $ -5n + 3 $ | $ -5(1)+3 = -2 $ | $ -5(2)+3 = -7 $ | $ -5(3)+3 = -12 $ | $ -5(4)+3 = -17 $ | $ -5(5)+3 = -22 $ |
| $ \frac{1}{2}n $ | $ \frac{1}{2}(1) = 0.5 $ | $ \frac{1}{2}(2) = 1 $ | $ \frac{1}{2}(3) = 1.5 $ | $ \frac{1}{2}(4) = 2 $ | $ \frac{1}{2}(5) = 2.5 $ |
| $ \frac{3}{4}n $ | $ \frac{3}{4}(1) = 0.75 $ | $ \frac{3}{4}(2) = 1.5 $ | $ \frac{3}{4}(3) = 2.25 $ | $ \frac{3}{4}(4) = 3 $ | $ \frac{3}{4}(5) = 3.75 $ |

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✔ Section C: Find the first four terms (n = 1 to 4)

We substitute $ n = 1, 2, 3, 4 $

1) $ 2n - 12 $:
$ 2(1)-12 = -10 $, $ 2(2)-12 = -8 $, $ 2(3)-12 = -6 $, $ 2(4)-12 = -4 $ → -10, -8, -6, -4

2) $ 6n - 10 $:
$ 6(1)-10 = -4 $, $ 6(2)-10 = 2 $, $ 6(3)-10 = 8 $, $ 6(4)-10 = 14 $ → -4, 2, 8, 14

3) $ n - 98 $:
$ 1-98 = -97 $, $ 2-98 = -96 $, $ 3-98 = -95 $, $ 4-98 = -94 $ → -97, -96, -95, -94

4) $ 3n + 7 $:
$ 3(1)+7 = 10 $, $ 3(2)+7 = 13 $, $ 3(3)+7 = 16 $, $ 3(4)+7 = 19 $ → 10, 13, 16, 19

5) $ 2.5n $:
$ 2.5(1) = 2.5 $, $ 2.5(2) = 5 $, $ 2.5(3) = 7.5 $, $ 2.5(4) = 10 $ → 2.5, 5, 7.5, 10

6) $ 4 - 2n $:
$ 4-2(1) = 2 $, $ 4-2(2) = 0 $, $ 4-2(3) = -2 $, $ 4-2(4) = -4 $ → 2, 0, -2, -4

7) $ 8 - n $:
$ 8-1 = 7 $, $ 8-2 = 6 $, $ 8-3 = 5 $, $ 8-4 = 4 $ → 7, 6, 5, 4

8) $ 0.5n + 3 $:
$ 0.5(1)+3 = 3.5 $, $ 0.5(2)+3 = 4 $, $ 0.5(3)+3 = 4.5 $, $ 0.5(4)+3 = 5 $ → 3.5, 4, 4.5, 5

9) $ 6 - 3n $:
$ 6-3(1) = 3 $, $ 6-3(2) = 0 $, $ 6-3(3) = -3 $, $ 6-3(4) = -6 $ → 3, 0, -3, -6

10) $ 6n $:
$ 6(1) = 6 $, $ 6(2) = 12 $, $ 6(3) = 18 $, $ 6(4) = 24 $ → 6, 12, 18, 24

11) $ \frac{n}{2} $:
$ \frac{1}{2} = 0.5 $, $ \frac{2}{2} = 1 $, $ \frac{3}{2} = 1.5 $, $ \frac{4}{2} = 2 $ → 0.5, 1, 1.5, 2

12) $ 8 - 5n $:
$ 8-5(1) = 3 $, $ 8-5(2) = -2 $, $ 8-5(3) = -7 $, $ 8-5(4) = -12 $ → 3, -2, -7, -12

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✔ Section D: Find the first three terms (n = 1, 2, 3)

1) $ 2(n+1) $:
$ 2(1+1) = 4 $, $ 2(2+1) = 6 $, $ 2(3+1) = 8 $ → 4, 6, 8

2) $ n^2 $:
$ 1^2 = 1 $, $ 2^2 = 4 $, $ 3^2 = 9 $ → 1, 4, 9

3) $ \frac{n+1}{n} $:
$ \frac{1+1}{1} = 2 $, $ \frac{2+1}{2} = 1.5 $, $ \frac{3+1}{3} = \frac{4}{3} \approx 1.33 $ → 2, 1.5, 1.33...

4) $ 3(n+5) $:
$ 3(1+5) = 18 $, $ 3(2+5) = 21 $, $ 3(3+5) = 24 $ → 18, 21, 24

5) $ n^2 + 1 $:
$ 1^2+1 = 2 $, $ 2^2+1 = 5 $, $ 3^2+1 = 10 $ → 2, 5, 10

6) $ \frac{3n}{2} $:
$ \frac{3(1)}{2} = 1.5 $, $ \frac{3(2)}{2} = 3 $, $ \frac{3(3)}{2} = 4.5 $ → 1.5, 3, 4.5

7) $ 10(n - 9) $:
$ 10(1-9) = -80 $, $ 10(2-9) = -70 $, $ 10(3-9) = -60 $ → -80, -70, -60

8) $ n^2 + 5 $:
$ 1^2+5 = 6 $, $ 2^2+5 = 9 $, $ 3^2+5 = 14 $ → 6, 9, 14

9) $ n(n+2) $:
$ 1(1+2) = 3 $, $ 2(2+2) = 8 $, $ 3(3+2) = 15 $ → 3, 8, 15

10) $ 25(1 - n) $:
$ 25(1-1) = 0 $, $ 25(1-2) = -25 $, $ 25(1-3) = -50 $ → 0, -25, -50

11) $ 2n^2 $:
$ 2(1)^2 = 2 $, $ 2(2)^2 = 8 $, $ 2(3)^2 = 18 $ → 2, 8, 18

12) $ n^3 $:
$ 1^3 = 1 $, $ 2^3 = 8 $, $ 3^3 = 27 $ → 1, 8, 27

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📌 Summary of Answers

#### Section A:
| Expression | 1 | 2 | 3 | 4 | 5 |
|----------|----|----|----|----|----|
| $ n+2 $ | 3 | 4 | 5 | 6 | 7 |
| $ n-7 $ | -6 | -5 | -4 | -3 | -2 |
| $ 4n $ | 4 | 8 | 12 | 16 | 20 |
| $ 3n+1 $ | 4 | 7 | 10 | 13 | 16 |
| $ 5n-8 $ | -3 | 2 | 7 | 12 | 17 |
| $ 2n+0.5 $ | 2.5 | 4.5 | 6.5 | 8.5 | 10.5 |
| $ 12n-1.5 $ | 10.5 | 22.5 | 34.5 | 46.5 | 58.5 |

#### Section B:
| Expression | 1 | 2 | 3 | 4 | 5 |
|----------|----|----|----|----|----|
| $ 2-n $ | 1 | 0 | -1 | -2 | -3 |
| $ 13-n $ | 12 | 11 | 10 | 9 | 8 |
| $ -3n $ | -3 | -6 | -9 | -12 | -15 |
| $ 1-2n $ | -1 | -3 | -5 | -7 | -9 |
| $ -5n+3 $ | -2 | -7 | -12 | -17 | -22 |
| $ \frac{1}{2}n $ | 0.5 | 1 | 1.5 | 2 | 2.5 |
| $ \frac{3}{4}n $ | 0.75 | 1.5 | 2.25 | 3 | 3.75 |

#### Section C:
1) -10, -8, -6, -4
2) -4, 2, 8, 14
3) -97, -96, -95, -94
4) 10, 13, 16, 19
5) 2.5, 5, 7.5, 10
6) 2, 0, -2, -4
7) 7, 6, 5, 4
8) 3.5, 4, 4.5, 5
9) 3, 0, -3, -6
10) 6, 12, 18, 24
11) 0.5, 1, 1.5, 2
12) 3, -2, -7, -12

#### Section D:
1) 4, 6, 8
2) 1, 4, 9
3) 2, 1.5, $ \frac{4}{3} $ (≈1.33)
4) 18, 21, 24
5) 2, 5, 10
6) 1.5, 3, 4.5
7) -80, -70, -60
8) 6, 9, 14
9) 3, 8, 15
10) 0, -25, -50
11) 2, 8, 18
12) 1, 8, 27

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✔ All sequences have been generated correctly by substituting $ n = 1, 2, 3, \ldots $ into the given formulas. This method is fundamental in understanding how sequences are built from their nth term formula.

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