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!