Respuesta :
We have the recursive sequence
[tex]\begin{gathered} a_1=40,a_n=\frac{1}{2}\cdot a_{n-1} \\ \end{gathered}[/tex]A. Calculating the first four terms of the sequence, we get:
[tex]\begin{gathered} a_1=40 \\ a_2=\frac{1}{2}\cdot40=20 \\ a_3=\frac{1}{2}\cdot a_2\rightarrow a_3=\frac{1}{2}\cdot20=10 \\ a_4=\frac{1}{2}\cdot a_3\rightarrow a_4=\frac{1}{2}\cdot10=5 \end{gathered}[/tex]This way,
[tex]\begin{gathered} a_1=40 \\ a_2=20 \\ a_3=10 \\ a_4=5 \end{gathered}[/tex]B. Let's calculate the 9th term:
[tex]\begin{gathered} a_5=\frac{1}{2}\cdot a_4\rightarrow a_5=\frac{1}{2}\cdot5=\frac{5}{2} \\ a_6=\frac{1}{2}\cdot a_5\rightarrow a_6=\frac{1}{2}\cdot\frac{5}{2}=\frac{5}{4} \\ a_7=\frac{1}{2}\cdot a_6=\frac{1}{2}\cdot\frac{5}{4}=\frac{5}{8} \\ a_8=\frac{1}{2}\cdot a_7=\frac{1}{2}\cdot\frac{5}{8}=\frac{5}{16} \\ a_9=\frac{1}{2}\cdot a_8\rightarrow a_9=\frac{1}{2}\cdot\frac{5}{16}=\frac{5}{32} \end{gathered}[/tex]This way, we get that:
[tex]a_9=\frac{5}{32}[/tex]Now, we need to compare
[tex]\begin{gathered} \frac{1}{10} \\ \text{and} \\ \frac{5}{32} \end{gathered}[/tex]To do so, we'll use their LCM: 160
[tex]\begin{gathered} \frac{1}{10}\rightarrow\frac{16}{160} \\ \text{and} \\ \frac{5}{32}\rightarrow\frac{25}{160} \end{gathered}[/tex]Therefore, we can conclude that
[tex]a_9>\frac{1}{10}[/tex](The term is larger than 1/10)
