the logarithm of each member of the proposed equation, we have A = la + nl (1 + r,)

By means of this last equation we determine how many years the principal a must remain at interest in order to amount to a sum A.

To illustrate this by an example, I shall suppose that it is required to find the time in which the original sum will be doubled, the rate of interest being 5 per cent.; we have

259. The following question is one of the most complicated, that we meet with relating to this subject. We suppose, that the lender during a number n of years, adds each year a new sum, to the amount of this year; it is required to find what will be the value of these several sums, together with the compound interest that may thence arise at the expiration of the term proposed. Let a, b, c, d, . . . . k, be the sums added the first, second, third, fourth, &c. years; the sum a remaining in the hands of the borrower during a number n of years, amounts to

a (1 + r)" ;

the sum b, which remains n

the sum c, which remains n

1 years only, becomes

b (1 + r)n−1,

2 years only, becomes

c (1 + r)n−2,

and so on; the last sum, k, which is employed only one year, becomes simply

we have, therefore,

k (1 + r);

A = a(1+r)" + b (1 + r)n−1 + c (1 + r)n−2 . +k(1+r).

....

By calculating the several terms of the second number separately, we obtain the value of A.

The operation is very much simplified when

a = b = c=d. . . . = k,

for in this case we have

r)n−2.

A = a (1 + r)” + a(1 + r)”~1 + a (1 + r)~~2 . . . . + a (1 + r) ; the second member of this equation forms a progression by quotients, of which the first term is a (1+r), the last term a (1+r)", the quotient + r, and the sun, consequently,

This equation gives rise also to four questions corresponding to those mentioned in connexion with the equation.

A = a (1 + r)”.

260. By reversing the case we have been considering, we may represent those annual sums, or sums due at stated intervals, called annuities; here the borrower discharges a debt with the interest due upon it, by different payments made at regular periods. These payments, made by the borrower before the debt in question is discharged, may be considered, as sums advanced to the lender toward the discharge of the debt, the value of which sums will depend upon the interval of time between the payment and the expiration of the annuity. Thus, if we represent each sum by a, the first payment, which will take place n1 years before the expiration of the term of the annuity, referred to this time, is worth a (1+r); the second referred to the same epoch, is worth only a (1+r)-2; the third, a (1 + r)”—3, and so on to the last, which amounts only to the value of a. But on the other hand, the sum lent being represented by A, will be worth in the hands of the borrower, after n years, A (1+r)", which must be equal to the amount of the several payments advanced by him to the lender; we have, therefore, A(1+r)" = a(1 + r)n−1 + a (1 + r)"−2+ a (1 + r)"—3... +a, or taking the sum of the progression, which constitutes the second member

a —

£î (1 + r)" = a [ (1 + r)" — 1]

an equation, in which we may take for the unknown quantity, successively, the quantity A, which I shall call the value of the annuity, because it is the sum, which it represents, the quantity a, which is the quota of the annuity, the quantity r, which is the rateof interest, and lastly, the quantity n, which denotes the term of“. the annuity. In order to find this last we must have recourse to logarithms. We first disengage (1+r)", which gives

a

(1 + r)" = a— Ar'

then taking the logarithms, we have

nl (1+r) = la—1(a—Ar),

261. To give an instance of the application of the above formulas, I shall take the following question;

To find what sum must be paid annually to cancel in 12 years a debt of 100 dolls. with the interest during that time, the rate of interest being 5 per cent.

In this example the quantities given are

A = 100, n = 12, r =

1

20'

and the annuity a is required to be found; resolving the equation

a

A (1 + r)n = a [ (1 + r)" − 1]

with reference to the letter a, we have

Ar (1+r)" α= (1 + r)" — 1°

The values of the letters, A, r, and n, are to be substituted in this expression; and it will be found most convenient in the first place to calculate, by the help of logarithms, the quantity (1 + r)”, which becomes (1)12; and

and, determining the values of this last expression either directly or by means of logarithms, we find

a = 11,2826;

an annuity of 11,28 dolls., therefore, is necessary to cancel in 12 years a debt of 100 dolls., the rate of interest being 5 per cent.

262. I am prevented from entering into further details on this. subject by the limits I have prescribed myself in this treatise; I will merely add, therefore, that in order to compare the values of different sums, as they concern the person, who pays or receives them, they must be reduced to the same epoch, that is, we must find what they would amount to when referred to the same date. A banker, for instance, owes a sum a payable in n years; as an equivalent he gives a note, the nominal value of which is represented by b, and which is payable in p years, the first sum at the

a

time the note is given, is worth only (1+r) because it must be

considered as the original value of a principal, which amounts to a at the expiration of n years; the sum b, for the same reason, is

represents, therefore, according as it is positive or negative, what the banker ought to give or receive by way of balance; if this balance is not to be paid until after a number of years denoted by 9, c representing its value at the time the exchange is made, it will amount at the expiration of this term to

c (1 + r);

so that it will be equivalent to

α

(1+r)n

b

(1 + r)x ) (1+r)2 = a (1+r)s—*—b (1 + r)s—p.

The several sums, a, b,

.....

k, in art. 259., were reduced to the time of the payment of the sum A, and in art. 260., each of the payments, as well as the sum A, was referred to the time, when the annuity was to cease.

Questions relating to Interest and Annuities.

1. A capital of 5000l. stands at 4 per cent. compound interest. What will it amount to in 40 years?

Ans. 24005.1037.

2. What will 3200l. amount to at 3 per cent in 80 years?

Ans. 34050.847.

3. How long must a capital a remain at the interest p to become as much as a capital a' at the interest p' for n years? log an log p' — log a

Ans.

log p

years.

4. How long must 36007. remain at 5 per cent. compound interest so that it may become as much as 5000l. at 4 per cent. for 12 years? Ans. 16 years, nearly. 5. What is the amount of capital which at the interest p for n years is of equal value with a capital a' for n' years at the interest p'? Ans. log a log a' + n'. log p' n log p. 6. What is the amount of a capital which stands at 4 per cent. that 15 years hence it may be equal in value with 4500l. at 6 per cent. for 9 years? Ans. 42211., nearly. 7. What is the rate of interest that a capital a in n years may be equal to a capital a' in n' years at the interest p'? log an' log p' - log a

Ans. log p =

n

8. How long must a capital stand at 4 per cent. compound interest that it may double itself; and how long that it may be tripled ?

Ans. It doubles itself in between 17 and 18 years, and triples itself in between 28 and 29 years.

9. An usurer lent a person 600l. and drew up for the amount a bond payable in 3 years bearing no interest. What did he take per cent. reckoning compound interest?

Ans. 10 per cent., nearly. 10. A capital of 8007. increased in the space of 6 years to 3600l. What did it gain per cent. ?

Ans. 28 per cent., nearly. 11. A person enjoys an annuity of 500l. for six years. How much ready money can a person give him for this annuity, calculating 31 per cent. ? Ans. 26641. 5s. 10d.

12. What is the present value of an annuity of 350l. assigned for 8 years at 4 per cent.? Ans. 23561. 9s. 2d., nearly.

13. A debt due at this present time amounting to 1200l. is to be discharged in seven yearly and equal payments. What is the amount of these payments if the interest be calculated at 4 per cent. ? Ans. 2001., nearly.

14. A person wishes to obtain an annuity of 2000l. for 345801. For how many years can this annuity be granted him, computing the interest at 4 per cent.? Ans. About 30 years.