Part A.
The inital salary is $58,000, then we have:
[tex]a_1=58000_{}[/tex]
Since we have an increase of 3% each year we know that the second year the salary would be:
[tex]\begin{gathered} a_2=1.03a_1 \\ a_2=1.03\cdot58000 \end{gathered}[/tex]
The third year the salary would be:
[tex]\begin{gathered} a_3=1.03a_2 \\ a_3=1.03(1.03)58000 \\ a_3=(1.03)^258000 \end{gathered}[/tex]
and so on for year 4 and 5.
Since the increase in salary is only the first five years we conclude that this can't be represented by a geometric series.
For the first five year we can calculate the salary using a geometric sequence with common ratio 1.03, then for the first five years the salary is given by
[tex]a_n=(1.03)^{n-1}_{}\cdot58000\text{ for }1\leq n\leq5[/tex]
The salary for the any subsequent year is given by:
[tex]a_n=(1.03)^4\cdot58000\text{ for }n>5[/tex]
Part B.
Since we are adding a certain quantity each year we conclude that this offer can be represetend by an algebraic series given by:
[tex]\begin{gathered} b_n=56000+(n-1)3000 \\ b_n=56000+3000n-3000 \\ b_n=3000n+53000 \end{gathered}[/tex]
Part C.
After five years the income for offer A is:
[tex]a_5=(1.03)^4\cdot58000=65279.51[/tex]
For offer B is:
[tex]b_5=3000(5)+53000=68000_{}[/tex]
Therefore after 5 years job offer B has a greater total income.
