Let's examine each sequence step-by-step to determine whether it is arithmetic or geometric and then find the formula for the [tex]\(n^{\text{th}}\)[/tex] term [tex]\(a_n\)[/tex].
### Sequence (a): 6, 30, 150, ...
1. Determine the type:
- To identify if the sequence is geometric, let's calculate the ratio between consecutive terms:
[tex]\[
\text{Ratio} = \frac{30}{6} = 5 \quad \text{and} \quad \frac{150}{30} = 5
\][/tex]
- The ratio between consecutive terms is constant, so the sequence is geometric with a common ratio [tex]\( r = 5 \)[/tex].
2. Find the formula for [tex]\(a_n\)[/tex]:
- The formula for the [tex]\(n^{\text{th}}\)[/tex] term [tex]\(a_n\)[/tex] of a geometric sequence is given by:
[tex]\[
a_n = a_1 \cdot r^{(n-1)}
\][/tex]
- Here, [tex]\(a_1 = 6\)[/tex] and [tex]\(r = 5\)[/tex]:
[tex]\[
a_n = 6 \cdot 5^{(n-1)}
\][/tex]
So the sequence is geometric and the [tex]\(n^{\text{th}}\)[/tex] term formula is:
[tex]\[
\boxed{a_n = 6 \cdot 5^{(n-1)}}
\][/tex]
### Sequence (b): 17, 21, 25, ...
1. Determine the type:
- To identify if the sequence is arithmetic, let's calculate the difference between consecutive terms:
[tex]\[
\text{Difference} = 21 - 17 = 4 \quad \text{and} \quad 25 - 21 = 4
\][/tex]
- The difference between consecutive terms is constant, so the sequence is arithmetic with a common difference [tex]\(d = 4\)[/tex].
2. Find the formula for [tex]\(a_n\)[/tex]:
- The formula for the [tex]\(n^{\text{th}}\)[/tex] term [tex]\(a_n\)[/tex] of an arithmetic sequence is given by:
[tex]\[
a_n = a_1 + (n - 1) \cdot d
\][/tex]
- Here, [tex]\(a_1 = 17\)[/tex] and [tex]\(d = 4\)[/tex]:
[tex]\[
a_n = 17 + (n - 1) \cdot 4
\][/tex]
So the sequence is arithmetic and the [tex]\(n^{\text{th}}\)[/tex] term formula is:
[tex]\[
\boxed{a_n = 17 + (n - 1) \cdot 4}
\][/tex]
Putting it all together:
\begin{tabular}{|c|c|c|}
\hline
Sequence & Type & [tex]\(n^{\text{th}}\)[/tex] term formula \\
\hline
(a) [tex]\(6, 30, 150, \ldots\)[/tex] & Geometric & [tex]\(a_n = 6 \cdot 5^{(n-1)}\)[/tex] \\
\hline
(b) [tex]\(17, 21, 25, \ldots\)[/tex] & Arithmetic & [tex]\(a_n = 17 + (n-1) \cdot 4\)[/tex] \\
\hline
\end{tabular}
