Geometric Series - Quite big qu.

Quite a big Question here so when im next on i will rep one of the responders!!

A company buys a new car for £12 000 at the start of one year. In a model, it is assumed
that each year the value of a car decreases by 25% of its value at the start of that year.

a) Show that the value of the car after four years is £3800 to 3 significant figures.
The company plans to buy one new car for £12 000 at the start of each subsequent year.

b) Using the same model, find the total value of all the cars the company will have bought
under this plan immediately after the purchase of the eighth car.

Reply 1

Coolraj

(a) from the question, we know that;
a = 12 000
n = 5
r = 0.75

substituting these values into the equation Un =arn-1 gives:
Un = 12000(0.75)^4
= 3796.87
= £3800 (3 s.f)
Q.E.D

(b) I'm not quiet sure what the question is asking to calculate...

Reply 2

danniella

Hello!

Second part of the question
After 1 year, company have 2 cars, which are worth
12,000x0.75 (original car) + 12,000 (the one they've just bought)

After 2 years, they have 3 cars. worth
12,000x0.75x0.75 + 12,000x0.75 + 12,000
(12,000x0.75^2 + 12,000x0.75 + 12,000)

After 3 years, they have 4 cars,worth
12,000x0.75^3 + 12,000x0.75^2 + 12,000x0.75 + 12,000
=12,000(0.75^3 + 0.75^2 + 0.75 + 1)

So when they have 8 cars, the cars are worth
12,000(0.75^7 + 0.75^6 + 0.75^5 + 0.75^4 + 0.75^3 + 0.75^2 + 0.75 +1)
= 43194.58

or if you don't fancy typing all that into your calculator, it is
12,000 x S8 for a geometric series where a=1 and r-0.75
= 12,000[1(1-0.75^8)/(1-0.75)]
=12,000[(1-/6561/65536)/0.25]
=12,000(0.89989/0.25)
= 43194.58

love danniella

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