Too ARITHMETIC. 



4. Call tlie bquare of the third number the fourth num- 

 ber. 



5. Divide the product of the second payment, and time 

 between the payments, by the product of the first pay- 

 ment and the rate, and call tlie quotient the lifth 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. 



I. There is lool. payable one year hence, and 105I. 

 payable 3 years hence ; what is the equated time, allow- 

 ing simple interest at 5 per cent, per annum ? 



100 



^i^ 



But ti'^ — k X ] is evidently greater than n^ — l?i — at >< — 



ar \ ar 



and therefore n* — l>t X — I — n"- — k—atX- — 



ar 



, or its equal 



t — X, must be a negative quantity ; and consequently x will be 

 greater than t, that is, the equated time will fall beyond the sec- 

 ond payment, which is absurd. The value of x, therefore, can- 



, a-\-hA-atr . a-^-b-^-atr 

 not be = -X — ? + -J — ! 



2ar 



, but must in all cases 



bc=''J±±fl:-1±i±^\ -±\ , which is the same as 

 2ar 2ar \ ar | 



the rule. 



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, an-d that there is no ambiguity i^ ^e 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, 



