Answer
a) $A_n=1200n+33,800$
b) $35,000; 36,200; 37,400; 38,600,39,800,41,000; 42,200; 43,400;$
Work Step by Step
The salary $\{A_n\}$ is a arithmetic sequence with the elements:
$$\begin{cases}
A_1=35,000\\
d=1200.
\end{cases}$$
a) We determine the $n$th element of the sequence which is the salary in the $n$th year:
$$A_n=A_1+(n-1)d=35,000+1200(n-1)=1200n+33,800.$$
b) Calculate the first $8$ terms of the sequence:
$$\begin{align*}
A_1&=35,000\\
A_2&=A_1+d=35,000+1200=36,200\\
A_3&=A_1+2d=35,000+2(1200)=37,400\\
A_4&=A_1+3d=35,000+3(1200)=38,600\\
A_5&=A_1+4d=35,000+4(1200)=39,800\\
A_6&=A_1+5d=35,000+5(1200)=41,000\\
A_7&=A_1+6d=35,000+6(1200)=42,200\\
A_8&=A_1+7d=35,000+7(1200)=43,400.
\end{align*}$$
