Question

The price of a new car is `RM80000`. It is given that the price of the car depreciates at a constant rate if 5% yearly. Calculate the minimum number of years required for the price of the car to drop to less than `RM45000`.?

USE FORMULA : `T_n = ar^(n-1)`

Answer 1

`color(blue)(13)`

Explanation:

`ar^(n-1)`

This is just the nth term of a geometric sequence.

If the car is depreciating at `5%` yearly, then we can say that the value of the car after each year is `95%` of its value the previous year.

Using this as our common ratio and expressing it in decimal form, and using `a` as our initial value, the nth term will be:

`80000(0.95)^(n-1)`

We need this to be less than 45000. First we solve for `n` equal to this amount:

`80000(0.95)^(n-1)=45000`

Dividing by `80000`

`(0.95)^(n-1)=45000/80000=45/80=9/16`

Taking natural logarithms of both sides:

`(n-1)ln(0.95)=ln(9/16)`

`n-1=ln(9/16)/ln(0.95)`

`n=ln(9/16)/ln(0.95)+1~~12.21714158`

This is for equality, and we also need an integer value. We therefore go to the nearest integer greater than this:

`n=13`

So:

`T_13=80000(0.95)^(13-1)=43228.80702`