Answer:9000
Step-by-step explanation:
Given
Annual Raise =$ 1000
Time t=9 years
Let a be the initial salary
salary is growing in Arithmetic Progression
here common difference is d=1000
therefore salary after 9 th year i.e. salary of 10 th year is
[tex]a_9=a+(10-1)d [/tex]
[tex]a_9=a+9\times 1000[/tex]
initial salary is a
difference in salary is [tex]=a_9-a[/tex]
[tex]=a+9\times 1000-a[/tex]
[tex]=9\times 1000=9000[/tex]
so he will be earning $ 9000 more than initial salary