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Let the capital be $1200, and the rate 10 p. c., or per Dollar. Then, as 118-11, the series becomes : 1200: 1200 X 1200 (1): 1200 X (1)3, &c., of which the : 10 first term a 1200, the quotient q=1, and any term w equals 1200 (11)".

=

a: a

Therefore, calling a the first term, and r the interest of a dollar, the quotient will be 1+r, and we shall have × (1+r): a × (1+r)2: a × (1+r)3...... (1+r)"−1.

Increase of Population.

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1200 X

570. What is here said of compound interest, applies, word for word, to the increase of population.

To make use of the Table (465) for resolving all problems relative to Compound Interest and the Increase of Population, we shall substitute 1+r for q. Thus, instead of the formulæ Lω La

n-1

w

n = 1+

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we shall, for the first, have

Lq

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We have placed at interest $1000000, at the rate of ten per cent. compound interest. What sum shall we touch at the end

of five years and a half?

The sum that we shall touch at the end of the fifth year, will equal aq5. The interest for the six remaining months will equal the half of the interest of aq5 during a year; that is to say, aq5X; therefore, the sum that we shall touch at the end of five years and a half will equal aq5 + aq5 × 10 X=1000000(11)3+1000000(1})3×¡× 1 = $1691035).

Problem 2.

We have touched $1610510 for a capital, placed five years since, the compound interest being one-tenth per dollar; what was this capital ?

Since we know the last term 1610510(w,) the number of terms 6(n), the quotient (9), and that a =

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we shall

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We have touched $16910351 for a capital, at interest for five years and a half, the compound interest being at ten per cent. What was this capital?

Let a be the capital, the sum touched at the end of the fifth year will be a X (11), the interest for the six remaining months will equal the half of the interest of a(1) during a year; that is to say, a(11)5 X6 X 1, or a(11) X 20 We shall then have a(¦¦)3 + a(¦¦)3 × 20 1 w; that is to say, a × [ (13)3 + (}})3 × 2%]=w; therefore,

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1691035+

( ¦ ¦ ) 3 + ( 11 ) 3 X 210
16910351

3221020161051

2000000

(1691035+)2000000_3382071 × 1000000

dollars, the required capital.

3382071

Problem 4.

= 1000000

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

n-1

1610510(), the number of terms 6(n), and that q =

1000000

10

จะ

a

we

shall have q=/161851811; therefore, the interest was one-tenth per dollar, or ten per 100.

Problem 5.

Having placed at interest $1000000 at 10 per cent., we have touched $1691035. We demand what time the sum was at interest?

Since we know the first term 1000000(a), the last term

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=621188

41393

; therefore n =

the

1

which is one-tenth per dollar,
But the number of terms n =
L(1691035+1-L1000000
L11− L10

1+

But the sum touched at the end of the fifth year is equal to 1000000(11), that is to say, 1610510; therefore, 1691035+1 -1610510; that is to say, 80525 is the interest of 1610510 during a time Ρ of the sixth year. But this interest is equal to 16105101 Xp; that is to say, 161051 Xp; therefore, 80525+ 161051 Xp80525+; therefore, p=

The sum was therefore at interest 51⁄2 years.

Problem 6.

161051

= 10

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

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

@=

30000000 (181) 100.

Operating by logarithms, we shall find for the logarithm of (183) 2,00432-2,00000; that is to say, 0,00432; and for the logarithm of (181,)100, 0,43200, which corresponds to the number 2704 Therefore w = 30000000 780 30000 2704 · 81120000.

1000

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Therefore, at the end of a hundred and one years, France will contain 81120000 inhabitants.

Problem 7.

Suppose that in France there are 30000000 of inhabitants, and that the population augments by one hundreth per year: in how many years will the population amount to 81120000 inhabitants?

According to the formula n=

we shall have

then, as L81120000

n=1+

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- L30000000=0,43200, and L101 L1000,00432, we shall have n = =1+8; 38299=1+ 100

101.

0,00432

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

Problem 8.

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

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1+0,00432; now the logarithm 0,00432 answers to the number 101; therefore r— -1+180=180. The population must therefore increase by 10 yearly.

1009

Annuities.

571. According to the proper import of the word, an Annuity is a yearly income. Revenues arising from Public Funds or Stocks, Donations, Pensions, Salaries, Rents, &c., the payments of some of which are usually made at shorter periods, are also called Annuities.

The claim which any one holds, in expectancy, to a revenue dependent on a future event, as the demise of one or more individuals, &c., is called an Annuity in Reversion.

572. Public Funds or Stocks are loans advanced to Government, for which interest is regularly paid, from revenues set apart for the purpose. This mode of levying taxes for the payment of interest is called the Funding System, and the loans constitute the National Debt.

The debts of Government differ from other contracts in this, that the public creditor can claim only his interest; he may, however, sell his stock; that is, he may transfer his claim, and thus obtain his capital, more or less, according to the price of stock, which fluctuates from a variety of causes.

573. In raising loans a Douceur is sometimes given by Government of an annuity for a limited time; this is called a Terminable Annuity; but the returns of the regular stocks on which the common interest is paid, are called Perpetual Annuities.

574. Peyrard, a talented French writer, to whom we are indebted for a great part of this section, says, "We call annuities equal sums, paid yearly, to extinguish in a given time a capital and its compound interest."

A person having borrowed $12000 for five years, at ten per cent, would acquit himself in five equal yearly payments: what sum must he pay at the end of each year ?

It is evident that, if there were no reimbursements, he would be indebted 12000 × (11)5 at the end of the fifth year.

But the borrower remits a sum, b, at the end of the first year, without which b would have become b× (11)4; he remits b at the end of the second year, without which 6 would have become b× (1)3; he remits b at the end of the third year, without which b would have become b× (1)2; he remits b at the end of the fourth year, without which 6 would have become b×11; and, lastly, at the end of the fifth year, he remits b: therefore bX (1)1 +b ×(11) 3 + b × ( } } ) 3 + b× (44)+b=12000 (11)5.

But the numbers b × (1¿)a +b × (11)3 +6 × (11)2 + b× (1) +b form a geometrical progression, the sum of b × (11) × 1 - b 18-1

which is equal to


b × (11) 5 - b

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;

that is to say,

therefore

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but

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1]

therefore

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$316534785 Answer.

12000 × (11)5 × 10

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 a XrX (1+r)"

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With these three formulæ we can resolve all cases that

Annuities present.

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