1) Notice that:
[tex]\begin{gathered} 3=\frac{30}{10}, \\ \frac{3}{10}=\frac{3}{10}, \\ \frac{3}{100}=\frac{\frac{3}{10}}{10}. \end{gathered}[/tex]
Therefore the recursive formula for the first sequence is:
[tex]\begin{gathered} a_1=30, \\ a_n=\frac{a_{n-1}}{10}\text{ for }n\geq2. \end{gathered}[/tex]
2) Notice that:
[tex]\begin{gathered} 11=14-3, \\ 8=11-3, \\ 5=11-3. \end{gathered}[/tex]
Therefore the recursive formula for the second sequence is:
[tex]\begin{gathered} a_1=14, \\ a_n=a_{n-1}-3\text{ for }n\geq2. \end{gathered}[/tex]
Answer:
Left sequence:
[tex]\begin{gathered} a_1=30, \\ a_n=\frac{a_{n-1}}{10}\text{ for }n\geq2. \end{gathered}[/tex]
Right sequence:
[tex]\begin{gathered} a_1=14, \\ a_n=a_{n-1}-3\text{ for }n\geq2. \end{gathered}[/tex]
