Question

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Answer 1

a) 5, 11, 17, 23, 29

b) 1, 4, 7, 10, 13

c) 9, 17, 25, 33, 41

d) 2, 11, 26, 47, 74

Explanation:

a) 6n - 1

if n = 1

6(1) - 1

6 - 1 = 5

if n =2

6(2) - 1

12 - 1 =11

if n = 3

6(3) - 1

18 - 1 = 17

if n = 4

6(4) - 1

24 - 1 = 23

if n = 5

6(5) - 1

30 - 1 = 29

b) 3n - 2

if n = 1

3(1) - 2

3 - 2 = 1

if n =2

3(2) - 2

6 - 2 = 4

if n = 3

3(3) - 2

9 - 2 = 7

if n = 4

3(4) - 2

12 - 2 = 10

if n = 5

3(5) - 2

15 - 2 = 13

Answer 2

To find the first five terms in each of the sequences with the given nth terms, you simply need to substitute n with the values 1 through 5. Here are the first five terms for each sequence:

a. 6n - 1

- When n = 1: 6(1) - 1 = 6 - 1 = 5

- When n = 2: 6(2) - 1 = 12 - 1 = 11

- When n = 3: 6(3) - 1 = 18 - 1 = 17

- When n = 4: 6(4) - 1 = 24 - 1 = 23

- When n = 5: 6(5) - 1 = 30 - 1 = 29

So, the first five terms of this sequence are 5, 11, 17, 23, and 29.

b. 3n - 2

- When n = 1: 3(1) - 2 = 3 - 2 = 1

- When n = 2: 3(2) - 2 = 6 - 2 = 4

- When n = 3: 3(3) - 2 = 9 - 2 = 7

- When n = 4: 3(4) - 2 = 12 - 2 = 10

- When n = 5: 3(5) - 2 = 15 - 2 = 13

So, the first five terms of this sequence are 1, 4, 7, 10, and 13.

c. 8n + 1

- When n = 1: 8(1) + 1 = 8 + 1 = 9

- When n = 2: 8(2) + 1 = 16 + 1 = 17

- When n = 3: 8(3) + 1 = 24 + 1 = 25

- When n = 4: 8(4) + 1 = 32 + 1 = 33

- When n = 5: 8(5) + 1 = 40 + 1 = 41

So, the first five terms of this sequence are 9, 17, 25, 33, and 41.

d. 3n² - 1

- When n = 1: 3(1²) - 1 = 3(1) - 1 = 3 - 1 = 2

- When n = 2: 3(2²) - 1 = 3(4) - 1 = 12 - 1 = 11

- When n = 3: 3(3²) - 1 = 3(9) - 1 = 27 - 1 = 26

- When n = 4: 3(4²) - 1 = 3(16) - 1 = 48 - 1 = 47

- When n = 5: 3(5²) - 1 = 3(25) - 1 = 75 - 1 = 74

So, the first five terms of this sequence are 2, 11, 26, 47, and 74.