Answer:
a_8 =47
The cost of maintenance for 2 years is $69000
Explanation:
a.) a_0 = 23 is the number of cars at the beginning of the first month. At the end of the first month a_1 = 26 cars. This is due to the addition of extra 3 cars.
The Arithmetic Progression equation will be used to solve this question.
T(n) = a + (n-1)×d
Where
T(n) is the number of cars after n months
a is the number ofr cars at the end bbn of the first month; 26
d is the monthly car increment; 3
Substituting a and d into the equation, the equation reduces to
T(n)= 23 + 3n
For number of cars after 8 month a_8;
T(8)= 23 + 3(8)
T(8) = 47
b.) The maintenance cost at the first month is $50× 23 cars= $1150, for the second month $50 × 26 cars= $1300, for the third month $50 × 29 cars = $1450. The monthly increment is $150.
Arithmetic Progression will also be used to solve this problem.
Sm = n/2 [2a + (n-1)×d]
Where
Sn is the sum of maintenance cost for months n
a is the maintenance cost of the first month; $1150
d is the monthly increment; $150
Inserting a and d into the formula,Sn reduces to
Sn= n/2[2150 + 150n]
Inputting 24months (2 years) as n on the equation above
Sn= 24/2[2150 + 150×24]
Sn= $69000
