Solving for the 10th term for each of the recursive sequence
First sequence
[tex]\begin{gathered} a_1=32 \\ a_{n+1}=-5+a_n \\ \\ \text{This can be converted to} \\ a_n=a_1+(n-1)(-5) \\ \\ \text{Substitute }n=10 \\ a_{10}=32+(10-1)(-5) \\ a_{10}=32+(9)(-5) \\ a_{10}=32-45 \\ a_{10}=-13 \end{gathered}[/tex]Second sequence
[tex]\begin{gathered} a_1=2048 \\ a_{n+1}=-\frac{1}{2}a_n \\ \\ \text{This can be converted to} \\ a_n=a_1\cdot\Big(-\frac{1}{2}\Big)^{n-1} \\ \\ \text{Substitute }n=10 \\ a_{10}=2048\cdot\Big(-\frac{1}{2}\Big)^{10-1} \\ a_{10}=2048\cdot\Big(-\frac{1}{2}\Big)^9 \\ a_{10}=-4 \end{gathered}[/tex]Third sequence
[tex]\begin{gathered} a_1=0.125 \\ a_{n+1}=2a_n \\ \\ \text{This can be converted to} \\ a_n=a_1\cdot2^{n-1} \\ \\ \text{Substitute }n=10 \\ a_{10}=0.125\cdot2^{10-1} \\ a_{10}=0.125\cdot2^9 \\ a_{10}=64 \end{gathered}[/tex]Fourth sequence
[tex]\begin{gathered} a_1=-7\frac{2}{3} \\ a_{n+1}=a_n+1\frac{2}{3} \\ \\ \text{This can be converted to} \\ a_n=a_1+(n-1)\Big(1\frac{2}{3}\Big) \\ \\ \text{Substitute }n=10 \\ a_{10}=-7\frac{2}{3}+(10-1)\Big(1\frac{2}{3}\Big) \\ a_{10}=\frac{-23}{3}+(9)\Big(\frac{5}{3}\Big) \\ a_{10}=-\frac{23}{3}+\frac{45}{3} \\ a_{10}=\frac{22}{3} \\ a_{10}=7\frac{1}{3} \end{gathered}[/tex]Arranging the formulas from least to greatest according to their 10th terms, we have the following:
First Sequence → Second Sequence → Fourth Sequence → Third Sequence
