Geometrical
Progression

A
series of numbers is in geometrical progression if each number in the series is
a constant multiple of its preceding number. As an example, Macdonald gives the
series 2, 8, 32, 128, 512, 2048, 8192, etc. so that the constant multiple is 4.
The notes jump immediately into finding the sum of series in geometrical
progression. When the general form of the sum is given by

where is the first term in the series, is the ratio between succeeding terms, is the number of terms in the sum and is value of the sum. Macdonald derives
the value for in the usual way. Multiply both sides of the
equation by , find the
difference , and solve for to get

In
the initial numerical example, , , and is not given. Up to the “etc.” in the series, in which case .

In arithmetic textbooks and cyphering books of this
era, the usual application of geometrical progressions is to the valuation of
annuities at compound interest. If interest is paid at a rate of per annum, then capital
of invested at the rate per annum will accumulate to at the end of one year and at the end of two years (interest is earned on
interest). By the same concept a value of payable at the end of one year is worth now and a value of payable at the end of two years is worth now. This can be generalized to the value of
an annuity that pays at the end of every year for years. It is worth in today’s money

which is a geometrical progression
with and . Macdonald
does not treat interest and annuities. In fact, no form of commercial
arithmetic appears in the notebook. The whole notebook is an academic exercise
rather than training to become adept at commercial arithmetic.

The
three examples that Macdonald does treat are given as word problems. They all
originate in Nicholas Pike’s *A New and
Complete System of Arithmetick*. George Baxter
picked, and slightly edited, the more interestingly worded examples among the
ones that were available. The first two problems are to find the sum of the
series; the third problem is to find the last term in the series.

1. A gentleman, whose
daughter was married on New Year’s day, gave her a
guinea, promising to triple his gift on the first day of the month in the year.
What is the ladies [sic] fortune?

Answer:

2. An ignorant fop
wanting to buy an elegant house, a wily gentleman told him he had one which he
would sell on these moderate terms, viz. that he should give him a cent for the
first door, 2 for the 2^{d}, 4 for the 3^{d} so on doubling at
every door of which were 36 in all. It’s a bargain cries the simpleton. What
did the house cost him?

Answer:
$

3. A merchant wanting
to buy a cargo of horses for the West Indies, a jockey told him he would take
all the trouble and expense on himself of collecting 30 horses for [the] voyage
if he would give him the value of the last horse at the rate of ½ for the
first, 2d for the 2^{d}, 4d for the 3^{d} so on. What was the
value of the last horse? [The problem and solution are stated in pounds,
shillings and pence. One penny is abbreviated 1d and one shilling is 1s. Under
the pre-decimal system in the United Kingdom £1 = 20s and 1s = 12d, i.e. 20
shillings to the pound and 12 pence to the shilling. With the problem stated in
pence, Macdonald expressed his answer in pounds, shillings, and pence.]

Answer:
£ s d