We are given an arithmetic sequence that has a common difference of 10 and the 18th term is 180.
We will use the recursive formula.
In an arithmetic sequence, we have that the nth term is given by:
![a_n=a_1+(n-1)d](https://img.qammunity.org/2023/formulas/mathematics/high-school/8ad7drcg9vuq9sminhqfaw7j3r5r4u1ij9.png)
Where:
![\begin{gathered} a_n=nth\text{ term} \\ a_1=\text{ first term} \\ d=\text{ common difference} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7ikg83wq88gsbjmoutvhkukfd237px7pol.png)
Since the 18th term is 180 we have that:
![a_(18)=180](https://img.qammunity.org/2023/formulas/mathematics/college/x052spsgnrarumr5l39fj48hdgyn470tt0.png)
This means that when we substitute the value of "n" by 18 the result is 180:
![\begin{gathered} a_(18)=a_1+(18-1)(10) \\ \\ 180=a_1+(17)(10) \\ \\ 180=a_1+170 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5ehie7x9u5v2b3802ifj3fqcj8w5zwhxr9.png)
Now, we solve for the first term:
![\begin{gathered} 180-170=a_1 \\ 10=a_1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ngvjlhq00q7mulyolk9zzzgjonliedw7x0.png)
Now, we can apply the formula again using the first term:
![a_n=10+(n-1)(10)](https://img.qammunity.org/2023/formulas/mathematics/college/d2j1qw2iojq9g4cd7jlprmtsk36d1gq70j.png)
Now, we substitute the value of "n = 16";
![\begin{gathered} a_(16)=10+(16-1)(10) \\ \\ a_(16)=10+(15)(10) \\ \\ a_(16)=10+150 \\ \\ a_(16)=160 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dgmvk51fa6hlsopwp2gk7h9tgh5vft870r.png)
Therefore, the 16th term is 160