Answer: [tex]a_1=3,\quad a_n=\bigg(\dfrac{1}{6}\bigg)a_{n-1}[/tex]
Step-by-step explanation:
[tex]\text{The explicit rule is in the format:}\ a_n=a_1(r)^{n-1}\\\text{where}\ a_1\ \text{is the first term, r is the common ratio, and n is the term}[/tex]
[tex]\text{So,}\ a_n=3\bigg(\dfrac{1}{6}\bigg)^{n-1}\ \text{means that}\ a_1=3\ \text{and r}=\dfrac{1}{6}[/tex]
[tex]\text{The recursive rule applies when given the previous term.}\\ \text{The recursive formula is:}\ a_n=r\cdot a_{n-1}\quad \text{where}\ a_{n-1}\ \text{is the previous term}[/tex]
[tex]\text{So, inserting the r-value from the explicit formula into the recursive}\\\text{formula, we get:}\ a_n=\bigg(\dfrac{1}{6}\bigg)a_{n-1}[/tex]
