1.
The nth term is given by,
[tex]\begin{gathered} f(n)=f(n-1)+7 \\ f\mleft(1\mright)=2 \end{gathered}[/tex]
The common difference can be calculated as,
[tex]\begin{gathered} f(1)=2 \\ f(2)=f(1)+7=9 \\ f(3)=f(2)+7=16 \\ d=\text{ 7} \end{gathered}[/tex]
Therefore the tenth term can be calculated as,
[tex]\begin{gathered} a_n=a_1+d(n-1) \\ a_{10}=a_1+7\times9=2+63=65 \end{gathered}[/tex]
2.
SImilarly,
[tex]\begin{gathered} f(1)=30 \\ f(n)=2(fn-1)-50 \\ f(2)=2\times30-50=10 \\ f(3)=2\times10-50=-30 \\ d=-20 \end{gathered}[/tex]
Thus, the tenth term is,
[tex]f(10)=30-20\times(9)=-150[/tex]
Thus, the answer is -150.
