2. A owes B 521. 7s. 6d. to be paid in 4 months, 8ol. 10s. to be paid in 3 months, and 761. 2s. 6d. to be paid in 5 months; what is the equated time to pay the whole ? Ans. 4 months, 8 days. 3. A owes B 240l. to be paid in 6 months, but in one month and a half pays him 60l. and in 4 months after that 8ol. more; how much longer than 6 months should B in equity defer the rest? Ans. 3 month I I I 4. A debt is to be paid as follows, viz. at 2 months, at 3 months, at 4 months, at 5 months, and the rest at 7 months; what is the equated time to pay the whole ? Ans. 4 months and 18 days. EQUATION OF PAYMENTS BY DECIMALS. Tawo debts being due at different times, to find the equated time to pay the whole. RULE.* 1. To the sum of both payments add the continual product of the first payment, the rate, or intcrest of 11. for one year, and the time between the payments, and call this the first number. 2. Multiply * No rule in arithmetic has been the occasion of so many disputes, as that of Equation of Payments. Almost every writer upon this subject has endeavoured to shew the fallacy of the methods made use of 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 2. Multiply twice the first payment by the rate, and call this the second number. 3. Divide The rule, given above, is the same as Mr. MALCOLM's, except that it is not encumbered 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 10 be due immediately, and another sum at the expiration of a certain given time forward, 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 before due; because, in that case, the gain and loss will be equal, and consequently neither party can be the loser. Now, to find such a time, let a = first payment, b = second, and t = time between the payments; r = rate, or interest of Il. for one year, and x = equated time after the first payment. Then arx = interest of a for x time, btr-brx and = discount of b for the time t-x. 1+tr-rx But 3. Divide the first number by the second, and call the quotient the third number. Then it is evident that n, or its equal is greater than 이를 m2, and therefore & will have two affirmative values, the 2 quantities n+n2 m2 and n-n2. 2 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 Then t-xt-n-n2. 2 2 m=t-n -n m L m have from the first of these equations t2-2tn=-bt-at X I 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. EXAMPLES. 1. There is 1ool. payable one year hence, and 1051. payable 3 years hence; what is the equated time, allowing simple interest at 5 per cent. per annum ? TOO But n2-btx is evidently greater than n2-bt-atx I I 2 ---ax, or its equal ar t-x, must be a negative quantity; and consequently x will be greater that, that is, the equated time will fall beyond the second payment, which is absurd. The value of x, therefore, can From this it appears, that the double sigh 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 shewn, that the directions, usually given for finding the equated time when there are more than two payments |