Answer:
a) Number of projects in the first year = 90
b) Earnings in the twelfth year = $116500
Total money earned in 12 years = $969000
Step-by-step explanation:
Given that:
Number of projects done in fourth year = 129
Number of projects done in tenth year = 207
There is a fixed increase every year.
a) To find:
Number of projects done in the first year.
This problem is nothing but a case of arithmetic progression.
Let the first term i.e. number of projects done in first year = 
Given that:

Formula for
term of an Arithmetic Progression is given as:

Where
will represent the number of projects increased every year.
and
is the year number.

Subtracting (2) from (1):

By equation (1):

<em>Number of projects in the first year = 90</em>
<em></em>
<em>b) </em>
Number of projects in the twelfth year =

Each project pays $500
Earnings in the twelfth year = 233
500 = $116500
Sum of an AP is given as:

It gives us the total number of projects done in 12 years = 1938
Total money earned in 12 years = 500
1938 = $969000