Find the first five terms in sequences with the following nth terms:

a. [tex]\(9n - 1\)[/tex]
b. [tex]\(4n - 2\)[/tex]
c. [tex]\(6n + 1\)[/tex]
d. [tex]\(2n^2 - 1\)[/tex]

a. The first five terms of [tex]\(9n - 1\)[/tex] are [tex]\(\square\)[/tex], [tex]\(\square\)[/tex], [tex]\(\square\)[/tex], [tex]\(\square\)[/tex], and [tex]\(\square\)[/tex]. (Simplify your answers. Use ascending order.)



Answer :

Sure! Let's find the first five terms for each of the given sequences step-by-step:

1. Sequence with nth term given by [tex]\( 9n - 1 \)[/tex]:
- For [tex]\( n = 1 \)[/tex], the term is [tex]\( 9(1) - 1 = 9 - 1 = 8 \)[/tex]
- For [tex]\( n = 2 \)[/tex], the term is [tex]\( 9(2) - 1 = 18 - 1 = 17 \)[/tex]
- For [tex]\( n = 3 \)[/tex], the term is [tex]\( 9(3) - 1 = 27 - 1 = 26 \)[/tex]
- For [tex]\( n = 4 \)[/tex], the term is [tex]\( 9(4) - 1 = 36 - 1 = 35 \)[/tex]
- For [tex]\( n = 5 \)[/tex], the term is [tex]\( 9(5) - 1 = 45 - 1 = 44 \)[/tex]

So, the first five terms of [tex]\( 9n - 1 \)[/tex] are [tex]\( 8 \)[/tex], [tex]\( 17 \)[/tex], [tex]\( 26 \)[/tex], [tex]\( 35 \)[/tex], and [tex]\( 44 \)[/tex].

2. Sequence with nth term given by [tex]\( 4n - 2 \)[/tex]:
- For [tex]\( n = 1 \)[/tex], the term is [tex]\( 4(1) - 2 = 4 - 2 = 2 \)[/tex]
- For [tex]\( n = 2 \)[/tex], the term is [tex]\( 4(2) - 2 = 8 - 2 = 6 \)[/tex]
- For [tex]\( n = 3 \)[/tex], the term is [tex]\( 4(3) - 2 = 12 - 2 = 10 \)[/tex]
- For [tex]\( n = 4 \)[/tex], the term is [tex]\( 4(4) - 2 = 16 - 2 = 14 \)[/tex]
- For [tex]\( n = 5 \)[/tex], the term is [tex]\( 4(5) - 2 = 20 - 2 = 18 \)[/tex]

So, the first five terms of [tex]\( 4n - 2 \)[/tex] are [tex]\( 2 \)[/tex], [tex]\( 6 \)[/tex], [tex]\( 10 \)[/tex], [tex]\( 14 \)[/tex], and [tex]\( 18 \)[/tex].

3. Sequence with nth term given by [tex]\( 6n + 1 \)[/tex]:
- For [tex]\( n = 1 \)[/tex], the term is [tex]\( 6(1) + 1 = 6 + 1 = 7 \)[/tex]
- For [tex]\( n = 2 \)[/tex], the term is [tex]\( 6(2) + 1 = 12 + 1 = 13 \)[/tex]
- For [tex]\( n = 3 \)[/tex], the term is [tex]\( 6(3) + 1 = 18 + 1 = 19 \)[/tex]
- For [tex]\( n = 4 \)[/tex], the term is [tex]\( 6(4) + 1 = 24 + 1 = 25 \)[/tex]
- For [tex]\( n = 5 \)[/tex], the term is [tex]\( 6(5) + 1 = 30 + 1 = 31 \)[/tex]

So, the first five terms of [tex]\( 6n + 1 \)[/tex] are [tex]\( 7 \)[/tex], [tex]\( 13 \)[/tex], [tex]\( 19 \)[/tex], [tex]\( 25 \)[/tex], and [tex]\( 31 \)[/tex].

4. Sequence with nth term given by [tex]\( 2n^2 - 1 \)[/tex]:
- For [tex]\( n = 1 \)[/tex], the term is [tex]\( 2(1^2) - 1 = 2(1) - 1 = 2 - 1 = 1 \)[/tex]
- For [tex]\( n = 2 \)[/tex], the term is [tex]\( 2(2^2) - 1 = 2(4) - 1 = 8 - 1 = 7 \)[/tex]
- For [tex]\( n = 3 \)[/tex], the term is [tex]\( 2(3^2) - 1 = 2(9) - 1 = 18 - 1 = 17 \)[/tex]
- For [tex]\( n = 4 \)[/tex], the term is [tex]\( 2(4^2) - 1 = 2(16) - 1 = 32 - 1 = 31 \)[/tex]
- For [tex]\( n = 5 \)[/tex], the term is [tex]\( 2(5^2) - 1 = 2(25) - 1 = 50 - 1 = 49 \)[/tex]

So, the first five terms of [tex]\( 2n^2 - 1 \)[/tex] are [tex]\( 1 \)[/tex], [tex]\( 7 \)[/tex], [tex]\( 17 \)[/tex], [tex]\( 31 \)[/tex], and [tex]\( 49 \)[/tex].

Therefore, the first five terms for the sequences are:

a. The first five terms of [tex]\( 9n - 1 \)[/tex] are [tex]\( 8, 17, 26, 35, \)[/tex] and [tex]\( 44 \)[/tex].

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