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 = [tex]a[/tex]
Given that:
[tex]a_4=129\\a_{10}=207[/tex]
Formula for [tex]n^{th}[/tex] term of an Arithmetic Progression is given as:
[tex]a_n=a+(n-1)d[/tex]
Where [tex]d[/tex] will represent the number of projects increased every year.
and [tex]n[/tex] is the year number.
[tex]a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)[/tex]
Subtracting (2) from (1):
[tex]78 = 6d\\\Rightarrow d =13[/tex]
By equation (1):
[tex]129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90[/tex]
Number of projects in the first year = 90
b)
Number of projects in the twelfth year =
[tex]a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233[/tex]
Each project pays $500
Earnings in the twelfth year = 233 [tex]\times[/tex] 500 = $116500
Sum of an AP is given as:
[tex]S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938[/tex]
It gives us the total number of projects done in 12 years = 1938
Total money earned in 12 years = 500 [tex]\times[/tex] 1938 = $969000
