2. A owes B 521. 75. 6d. to be paid in 4 months, 8ol. 1os. to be paid in 3 months, and 761. 25. 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 2401. to be paid in 6 months, but in one month and a half pays him 601. and in 4 months after that sol. more ; how much longer than 6 months should B in equity defer the rest ? Ans. 31'- months 4. A debt is to be paid as follows, viz. Å at 2 months, 13 중 at 3 months, ğ at 4 months, ğ 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. Two 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 interest of il. for bne year, and the time between the payments, and call this the first number. 2. Multiply so many * No rule in arithmetic has been the occasion of disputes, as that of Equation of Paymchis. 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, des rived 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 io be due immediately, and another sum at the expiration of a certain given time forward, and it is proposed to find a time to ray 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 tims. 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 be. fore 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 i = time between the payments ; r = rate, or interest of 1!. for one year, and x = equated time after the first payment. Then arx = interest of a for x-time, and btr-bra Inoftr-rx = discount of b for the time tmex, But 3. Divide the first number by the second, and call the quotient the third number. 4. Call btr-orx But arx = by the question, from which equation x It-tr-rx 를 is found atb-tatr a+b+atr bt Zar 2ar ar Then it is evident that n, or its equal 777|* is greater than and therefore x will have two affirmative values, the 72 m7 2 m and 12-1 2 m quantities ntn? -n? being both positive. But only one of those values will answer the conditions of the question ; and, in all cases of this problem, & will be = n For suppose the contrary, and let.x = n +12=m*. n?. 2 m ;-2/ --m*=n2 +*24n m, we shall 2 ar ar 4. Call the square of the third number the fourth nuanber. 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 ioot. payable one year henice, and 1051. payable 3 years hence ; what is the equated time, allowing simple interest at 5 per cent. per annum ? 100 t_x, must be a negatire quantity ; and consequently will be greater thanz , , 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 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 shewn, that the directions, usually given for finding the equated time when there are more than two payments |