Review question

Ref: R9375

## Solution

A colour television set is hired for $£80$ in the first year. In each subsequent year the charge is four-fifth of the charge in the previous year. Write down the first four terms of a geometric progression of payments made in successive years, without working them out individually. Calculate the total paid over 8 years by the hirer to the nearest $£$.

For simplicity we drop all units as we work and keep in mind that costs are in $£$ and times in years.

The first four payments are $80, 80 \times \dfrac{4}{5}, 80 \times \left(\dfrac{4}{5}\right)^2, 80 \times \left(\dfrac{4}{5}\right)^3.$

These form the first four terms of a geometric progression, with $a = 80,$ and $r = \dfrac{4}{5}.$

The sum of the first $n$ terms of a geometric progression is $S_n = \dfrac{a(1-r^n)}{1-r}$.

We need $S_8 = \dfrac{80(1-0.8^8)}{1-0.8} = £333$ (to the nearest pound).

If instead he had borrowed $£200$ to buy the set outright and this debt had accumulated at $10\%$ compound interest, what total sum to the nearest $£$ would be owing after 8 years?

At the start, he owes $200$. After one year, he owes $200 \times 1.1$ (an increase of $10$ per cent). After $n$ years, he owes $200 \times 1.1^n.$

Thus after eight years, he owes $200 \times 1.1^8 = £429$ (to the nearest pound).

The question doesn’t tell us that the hirer owns the set after eight years, so which option is the best value is moot!