Answer:
Options A and C
Explanation:
We want to find out which arithmetic sequence(s) contain the term 34.
For an arithmetic sequence to contain the term, 34, the corresponding n-value must be an integer.
Option A
Set tn = 34
[tex]\begin{gathered} t_n=6+(n-1)4 \\ 34=6+(n-1)4 \end{gathered}[/tex]
Solve for n:
[tex]\begin{gathered} 34-6=4n-4 \\ 28=4(n-1) \\ n-1=\frac{28}{4}=7 \\ n-1=7 \\ n=7+1 \\ n=8 \end{gathered}[/tex]
The 8th term of this sequence is 34.
Option B
[tex]\begin{gathered} t_n=3n-1 \\ 34=3n-1 \\ 34+1=3n \\ 35=3n \\ n=\frac{35}{3}=11\frac{2}{3} \end{gathered}[/tex]
A sequence cannot have a decimal nth term, therefore, the sequence does not contain 34.
Option C
T1 = 12, d=5.5
[tex]\begin{gathered} 12+5.5(n-1)=34 \\ 5.5(n-1)=34-12 \\ 5.5(n-1)=22 \\ n-1=\frac{22}{5.5} \\ n=4+1 \\ n=5 \end{gathered}[/tex]
The 5th term of this sequence is 34, therefore, it contains the term 34.
Option D
3,7,11,...
[tex]\begin{gathered} t_1=3 \\ d=7-3=4 \end{gathered}[/tex]
Using the nth term of an arithmetic sequence formula:
[tex]\begin{gathered} t_n=t_1+(n-1)d \\ 34=3+4(n-1) \\ 34-3=4(n-1) \\ 31=4(n-1) \\ n-1=\frac{31}{4} \\ n-1=7\frac{3}{4} \\ n=8\frac{3}{4} \end{gathered}[/tex]
A sequence cannot have a decimal nth term, therefore, the sequence does not contain 34.
The sequences in Options A and C contain the term 34.
