Question

Please answer both i will give brainliest and thanks!!!! please quickly!!! hurry pleaseeee!!!

Answer 1

1.

The nth term is given by,

$\begin{gathered} f(n)=f(n-1)+7 \\ f\mleft(1\mright)=2 \end{gathered}$

The common difference can be calculated as,

$\begin{gathered} f(1)=2 \\ f(2)=f(1)+7=9 \\ f(3)=f(2)+7=16 \\ d=\text{ 7} \end{gathered}$

Therefore the tenth term can be calculated as,

$\begin{gathered} a_n=a_1+d(n-1) \\ a_{10}=a_1+7\times9=2+63=65 \end{gathered}$

2.

SImilarly,

$\begin{gathered} f(1)=30 \\ f(n)=2(fn-1)-50 \\ f(2)=2\times30-50=10 \\ f(3)=2\times10-50=-30 \\ d=-20 \end{gathered}$

Thus, the tenth term is,

$f(10)=30-20\times(9)=-150$

Thus, the answer is -150.