Question

Colleen's station wagon is depreciating at a rate of 9% per year. She paid $24,500 for it in 2002. What will the car be worth in 2008 to the nearest hundred dollars?

Answer 1

We have a geometric sequence :

`u_1 = 24500`
`u_(n+1) = u_n - u_n * 9/100 = u_n * 91/100`

`=> u_n = u_1 * r^(n-1)`, where `r = 91/100` and `u_n` gives you the worth of the car in the year `n`.

`=> u_n = 24500 * (91/100)^(n-1)`

`n = 1 = year 2002 => n = 7 = year 2008`

`=> u_7 = 24500 * (9/100)^6 ~~ 13900 $` in 2008.