We have put to interest $1000000, and have touched $1610510 at the end of five years. What was the compound interest?

Since we know the first term 1000000(a,) the last term 1610510(,) the number of the terms 6(n,) and that

therefore the interest was a tenth per dollar, or ten per 100.

PROBLEM V.

Having placed at interest $1000000 at ten per cent. we have touched $16910351. We demand what time the sum was at interest?

Since we know the first term 1000000(a,) the last term 1691035(,) and the interest which is a tenth per dollar; the quotient q will then be . But the number of the terms

But the sum touched at the end of the fifth year is equal to

1000000(+)5,

that is to say, 161050;

therefore

1691035+1610510,

that is to say, 80525 is the interest of 1610510 during a time p of the sixth year. But this interest is equal to

I suppose, in the three following problems, that the population in France augments by a hundredth per year, that is to say, that a family composed of one hundred persons to-day will contain a hundred and one at the end of one year. This is not a vague supposition; the last lists prove that the population of France, for the last twenty-five years, in spite of the cruel wars, is considerably augmented. This increase has two principal causes: the suppression of monasteries, and the discovery of vaccination.

If a long peace reigned in France, if vaccination were generally practised, and finally, if celibacy had ceased for ever to be placed among the number of the virtues, it is beyond a doubt that there would arrive an epoch when France would no longer be able to nourish its inhabitants; and this is proved by the three following problems.

PROBLEM VI.

Suppose that there are 30000000 of inhabitants in France, and that the population augments by a hundredth per year. According to these two suppositions, how many inhabitants will there be in France at the end of a hundred and one years?

Agreeably to the formula =aq"--1; we shall have'

@=30000000×(101) 10°.

Operating by logarithms, we shall find for the logarithm of (18) 2,00432-2,00000,

w=30000000 278=30000 × 2704=81120000. Therefore, at the end of a hundred and one years, France will contain 81120000 inhabitants.

PROBLEM VII.

Suppose that in France there are 30000000 of inhabitants, that the population augments by a hundredth per year. In how many years will the population amount to 81120000 inhabitants?

since L.81120000-L.30000000=0,43200,
and since L.101-L.100=0,00432.

0,43200

we shall have n=1+

=1+100=101.

0,00432

Therefore in a hundred and one years the population of France will be 81120000.

PROBLEM VIII.

There are 30000000 inhabitants in France; in what proportion must the population augment so that there may be 81120000 at the end of a hundred and one years s?

According to the formula,

now the logarithm 0,00432 answers to the number 1;

therefore r-1+10=.

The population must therefore augment by a hundredth yearly.

OF ANNUITIES.

We call annuities equal sums which are paid each year to extinguish, in a given time, a capital and its compound in

terest.

A person has borrowed $12000 for five years, at the rate of ten per cent.; he would acquit himself in five equal payments made at the end of each year. What is the sum that he will have to pay at the end of each year?

It is evident that, if there were no reimbursements, he would be indebted 12000×(1) at the end of the fifth year.

But the borrower remits a sum b, at the end of the first year, without which 6 would have become b×({}); he remits b at the end of the second year, without which 6 would have become bx(); he remits b at the end of the third year, without which 6 would have become b×(11); he remits b at the end of the fourth year, without which 6 would have become bx, and lastly, he remits b at the end of the fifth year: therefore

2

b. × ( { ↓) 1 +b × ( ¦¦)3+b× (}})2+b×(}})+b=12000(¦¦)3. But the numbers

b× ( } } ) • +b × (}})3+b×(}})2+b×{}+b

form a geometrical progression, the sum of which is equal to b × ( } 1 ) 4 × 1 1 - b

It will be the same whatever be the sum lent, the interest and the number of years.

Calling a the sum lent, n the number of years, r the interest, and b the sum which must be remitted at the end of each year, we shall have

With these three formula we can resolve all the cases that

annuities present.

Let it be proposed, for example, to find what sum we must give yearly, to extinguish, in three years, a debt of $500, the interest being at ten per cent.

Substituting these numbers in the first formula, we shall find b=$201+37. Therefore, $2013, or $201,057+ is the yearly payment.

If the interest were at 7 per cent., we should find b=$190,526-. This last answer being somewhat too great, we annex the sign

CONJOINED PROPORTION.

This rule is thus named, because it serves to find the relation of one number to another by means of several proportions.

Let a be the pound of Venice, b that of Lyons, c that of Rouen, d that of Toulouse, e that of Geneva, and let

100a 70b
1206=100c

80c 100d

100d= 74e;

It is required to find the relation of the pound of Venice a with the pound of Geneva e. The above equations furnish the following proportions: