EQUATION OF PAYMENTS BY DECIMALS. Two debts being due at different times, to Jind the equated time to pay the whole. RULE.* 1. To the sum of both payments add die continual product of the first payment, the rate, or interest of ll. for one year, and the time between the payments, and call this the first number. * No rule in Arithmetic has been the occasion of so many disputes, as that of Equation of Payments. Almost every writer on this subject has endeavoured to show the fallacy of the methods, used by other authors, and to substitute a new one in their stead. But the only true rule seems to be that of Mr. Malcolm, or one similar to it in its essential principles, derived from the consideration of interest and discount. The rule, given above, is the same as Mr. Malcolm's, except that it is not incumbered with the time before any payment is due, that being no necessary part of the operation. Demonstration Of The Rule. Suppose a sum of money to be due immediately, and another sum at the expiration of a certain given time, and it is proposed to find a time to pay the whole at once, so that neither party shall sustain loss. Now it is plain, that the equated time must fall between those of the two payments; and that what is got by keeping the first debt after it is due, should be equal to what is lost by paying the second debt before it is due. But the gain, arising from the keeping of a sum of money after it is due, is evidently equal to the interest of the debt for that time. And the loss, which is sustained by the paying of a sum of money before it is due, is evidently equal to the discount of the debt for that time. Therefore it is obvious, that the debtor must retain the sum immediately due, or the first payment, till its interest shall be equal to the discount of the second sum for the time it is paid Vol. I. Y 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 Uue; 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 t = time between the payments; r = rate, or interest of 11. for one year, and x = equated time after the first payment. Then arx =interest of a for x time, Then it is evident, that n, or its equal nis greater than n, nm, and therefore will have two affirmative values, the 2 2 2 quantities n+n-m 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 2 . For suppose the contrary, and let x = Then tx=t—n—n 1 n+n2. 1 t2-2tn+n22-n2 —m | 2 — n2 1 1 ar Now, since a+b+atr×; =n, and btx- =m, we shall 2ar have from the first of these equations t2-2tn= and consequently t-x-n2-bt-at X 1 ar But n2-6x is evidently greater than n2-bi-atx 1 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. ar 1 ar and therefore n2-bt×· -n2 bi-ai ×· , or its equal tx, must be a negative quantity; and consequently x will be greater than I, 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 compound interest, and is found in the same manner as the former. EXAMPLES. 1. There is 1001. payable one year hence, and 1051. payable 3 years hence; what is the equated time, allowing simple interest at 5 per cent. per annum ? 100 100 2 2. Suppose 4001. are to be paid at the end of 2 years, and 21001. 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 3001. are to be paid at one year's end, and 3001. more at the end of 1 year; it is required to find the it at one payment, 5 per cent. simple interest beAns. 1 248437 year. time to pay ing allowed. 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 3201. 10s. for 4 years, at 5 per cent. per annum, compound interest? 389 11 41 whole amount, or the answer required. The reason of this rule is evident from the definition, apl the principles of simple interest. |