A Silicon Valley billionaire purchases 3 new cars for his collection at the end of every month. Let a_n denote the number of cars he has after n months. Let a_0 = 23.a) What is a_8 ?b) If he pays $50 each month to have a car maintained, how much does he pay for maintenance after 2 years? No need to calculate the actual number. Instead give a closed form (without the summation) mathematical expression for the number. Note that he purchases the new car at the end of each month, so during the first month, he is only maintaining 23 cars.

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

tallinn tallinn · Answer 2 · 2022-01-11T15:53:14+01:00

The correct answer is: a_8 =47

When The cost of maintenance for 2 years is $69000

a.) when a_0 = 23 is the number of cars at the beginning of the 1 month. and then At the end of the first month a_1 = 26 cars. an addition that of extra 3 cars.

Therefore The Arithmetic Progression equation will be used are:

That is T(n) = a + (n-1)×d

Where that T(n) is the number of cars after that n months

So that "a is the number of cars at the end bbn of the first month; 26 d is the monthly car increment that is 3

When Substituting a and d into the equation, the equation will be reduced to

That is T(n)= 23 + 3n

After that For number of cars after 8 month a_8;

which is T(8)= 23 + 3(8)

T(8) = 47

b.) Solution of second is that The maintenance of the cost at the first month is $50× 23 cars= $1150, for the second month is $50 × 26 cars= $1300, for the third month is $50 × 29 cars = $1450. The monthly increment is $150.

After that Arithmetic Progression will also be used to solve this problem are:

That is Sm = n/2 [2a + (n-1)×d]

Where that, Sn is the sum of maintenance which cost for months is n

When 'a' is the maintenance cost of the first month; $1150
Also that 'd' is the monthly increment; $150
After that Inserting a and d into the formula, Sn reduces to
That is Sn= n/2[2150 + 150n]
When they are Inputting 24months (2 years) as n on the equation above
Sn= 24/2[2150 + 150×24]
so, that solution is Sn= $69000

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