2. Multiply twice the first payment by the rate, and call this the second payment.

3. Divide the first number by the second, and call the quotient the third number.

before due; because, in that case, the gain and the loss will be equal, and consequently neither party can be loser.

Now to find such a time, let a = first payment, b = second, and = time between the payments; r = rate, or interest of 11. for one year, and x = equated time after the first payment. Then are interest of a for x time,

Then it is evident, that n, or its equal nis greater than

and therefore a will have two affirmative values, the

quantities n+n'

and n-n-m being both positive.

But only one of those values will answer the conditions of the question; and, in all cases of this problem, x will be n n2 — m | } .

For suppose the contrary, and let _x = n+n2. -772

have from the first of these equations t2—2tn=—bi—a

and consequently tx-n2-bt-atx

1

ar

-btx

ur

But n2-bt× is evidently greater than n3—bi-a

ar

4. Call the square of the third number the fourth number. 5. Divide the product of the second payment, and time between the payments, by the product of the first payment and the rate, and call the quotient the fifth number.

6. From the fourth number take the fifth, and call the square root of the difference the sixth number.

7. Then the difference of the third and sixth numbers is the equated time, after the first payment is due.

1

and therefore n2 btx 1 -n2-bt—at X- , or its equal t—x,

ar

a

must be a negative quantity; and consequently x will be great. er than , that is, the equated time will fall beyond the second payment, which is absurd. The value of x therefore cannot

From this it appears, that the double sign, made use of by Mr. MALCOLM, and every author since, who has given his method, cannot obtain, and that there is no ambiguity in the problem.

In like manner it might be shown, that the directions, usually given for finding the equated time, when there are more than two payments, will not agree with the hypothesis; but this may be easily seen by working an example at large, and examining the truth of the conclusion.

The equated time for any number of payments may be readily found when the question is proposed in numbers, but it would not be easy to give algebraic theorems for those cases, on account of the variation of the debts and times, and the difficulty of finding between which of the payments the equated time would happen.

Supposing r to be the amount of 11. for one year, and the othlog. art +b

rem for the equated time of any two payments, reckoning com pound interest, and is found in the same manner as the former.

180

EXAMPLES.

1. There is 1001. payable one ycar hence, and 1051. payable 3 years hence; what is the equated time, allowing simple interest at 5 per cent. per annum ?

100

100

'05

2

2. Suppose 400l. are to be paid at the end of 2 years, and 2100l. at the end of 8 years; what is the equated time for one payment, reckoning 5 per cent. simple interest?

Ans. 7 years.

3. Suppose 300l. are to be paid at one year's end, and 3001. more at the end of 1 year; it is required to find the time to pay it at one payment, 5 per cent. simple interest be ing allowed. Ans. 1 248437 year.

COMPOUND INTEREST.

COMPOUND INTEREST is that, which arises from the principal and interest taken together, as it becomes due, at the end of each stated time of payment.

RULE.*

1. Find the amount of the given principal for the time of the first payment by simple interest.

2. Consider this amount as the principal for the second payment, whose amount calculate as before, and so on through all the payments to the last, still accounting the last amount as the principal for the next payment.

EXAMPLES.

1. What is the amount of 320l. 10s. for 4 years, at 5 per cent. per annum, compound interest?

389

11 44 whole amount, or the answer required.

The reason of this rule is evident from the definition, and the principles of simple interest.

2. What is the compound interest of 7601. 10s. forborn 4 years, at 4 per cent. ? Ans. 1291. 3s. 6d. 3. What is the compound interest of 4101. forborn for 21 years, at 41 per cent. per annum; interest payable halfyearly? Ans. 481. 4s. 11 d.

4. Find the several amounts of 501. payable yearly, halfyearly and quarterly, being forborn 5 years, at 5 per cent. per annum, compound interest.

Ans. 651. 169. 31d. 641. and 641. 1s. 94d.

COMPOUND INTEREST BY DECIMALS.

RULE.*

1. Find the amount of 11. for one year at the given rate per cent.

=

* DEMONSTRATION. Let r amount of 11 for one year, and p principal or given sum; then since r is the amount of 11. for one year, r2 will be its amount for two years, r3 for 3 years, and so on; for when the rate and time are the same, all principal sums are necessarily as their amounts; and consequently as r is the principal for the second year, it will be as 1:r::r amount for the second year, or principal for the third; and again, as 1 :r :: r2 : r3 amount for the third year, or principal for the fourth, and so on to any number of years. And if the number of years be denoted by t, the amount of 11. for years will be rt. Hence it will appear, that the amount of any other principal sum for years is pr; for as 1 :::: pr, the same as in the rule.

=

=

If the rate of interest be determined to any other time than a year, as, 1, &c. the rule is the same, and then t will represent that stated time.

Let

r = amount of 11, for one year at the given rate per

Then the following theorems will exhibit, the solutions of all the cases in compound interest.