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The formula for an arithmetic series is shown below, where [tex]$n = 1, 2, 3, \ldots$[/tex]:

[tex]$f(n+1) = f(n) + 13$[/tex]

If [tex]$f(1) = 3$[/tex], what are the fourth, fifth, and sixth terms in the sequence?

A. [tex]$25, 28, 31$[/tex]
B. [tex]$29, 42, 55$[/tex]
C. [tex]$42, 55, 68$[/tex]
D. [tex]$42, 68, 94$[/tex]

Answer :

Let's solve the problem of finding the fourth, fifth, and sixth terms in the given arithmetic sequence.

We know that this is an arithmetic sequence where each term is obtained by adding a fixed number, called the common difference, to the previous term. The sequence is defined as follows:

- Start with the first term: [tex]\( f(1) = 3 \)[/tex]
- The common difference (the amount added to each term to get the next term) is [tex]\( 13 \)[/tex].

The formula for the [tex]\( n \)[/tex]-th term of an arithmetic sequence is:
[tex]\[ f(n) = f(1) + (n-1) \times d \][/tex]
where [tex]\( d \)[/tex] is the common difference.

Now, let's find the terms:

1. Fourth Term:
- To find [tex]\( f(4) \)[/tex], use the formula:
[tex]\[
f(4) = f(1) + 3 \times 13 = 3 + 39 = 42
\][/tex]

2. Fifth Term:
- To find [tex]\( f(5) \)[/tex], use the formula:
[tex]\[
f(5) = f(1) + 4 \times 13 = 3 + 52 = 55
\][/tex]

3. Sixth Term:
- To find [tex]\( f(6) \)[/tex], use the formula:
[tex]\[
f(6) = f(1) + 5 \times 13 = 3 + 65 = 68
\][/tex]

The fourth, fifth, and sixth terms in the sequence are [tex]\( 42, 55, \)[/tex] and [tex]\( 68 \)[/tex], respectively.

Therefore, the correct answer is C: [tex]\( 42, 55, 68 \)[/tex].

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