I ( 153,06 51,86 Therefore ifr+=a IXC 21:14:= 4 x 2 @?:6 3brv+2bretbeerbaa 4rort3rret3r1= a a a 2 in Numb. 5349160 + 8729 € 3 in Numb. 6118400 + 59206 + 7466 4 in Numb. 7 64000 + 4800 • + 120 e +6+7 81531560 + 194494 + 19411=560783 8 - 531560 9 194496 + 1946. = 29223 9 • 194 10 100,2+ei= 153,06 =D D 10 1114 100,2+2 40,0T 101,2 41,5=rtosa 2d Divisor 101,7) 50,85 1,01 Os new r=41,5 for a second Operation, which being duly involved, &c. will be found more than just. Therefore irea 2 croca Then 3 | bor - 2 brotbeesbaa 41rrrirret3r1= a a a These being turned into Numbers, &c. as above, they will be 20037,75 - 198,5 !1 = 390,375, which being divided by 198,5 the Co-efficient of « , will become 100,946eee 1,966624, &c. = D. Operation 100,946 OI 41,5000000 =1 aft Divifor 100,936) 1,96624 0194847 = ,009 1,00936 41,4805153=r-154 2d Divisor 100,927) 957264 908343 Here I proceed by 489210 plain Division without 403708 forming new Divisors. 855020 807416 { .72332 &c. li 2 Let Let the laft Equation in the Enigma, Chap. g. be here proposed for a Solution. Viz. a a a atbaaacaa-da=G; b=2, c= 288,"d=506, and G = 1513, Quære a. By Tryals it will be found, that the next nearest r = 20, being something more than just. Therefore I'm a I xd 2dr de=da zcaa IQ 41 51144rrre torree=aa a a These being turned into Numbers, and those duly collected, according as the Signs of the Equation direct, they will become 50680 - 22374 et 2232e651513, which being all divided by 2232 the Co-ëfficient of ee, will be 10 e 11=22= D. By what hath been already done about the Solution of these few Equations (being carefully observed) I presume the Learnec will eatily conceive how to proceed in the Solution of all Kinds of Equations, be they never so high, or adfected; therefore I shall not here propore many various Examples, but only take them as they fall in Course, when I come to the next Part, wherein you will (perhaps) find fuch Equations with their Solutions as are not common. C H A P. XI. Of Simple Jnterett, annutties, or Penfions, &c. I NTEREST, or the Ufe paid for the Loan of Money, is either Simple; or Compound. Sect. 1. Of Simple Intereft. SIMPLE Intereft, is that which is paid for the Loan of any or , Rate per cent : agreed on between the Borrower and the Lender; which, according the late Laws of England, ought to be fix Pounds for the Use of 100 % for one Year, and twelve Pounds for the Use of 100 l, for two Years: and fo on for a greater, or lefser Sum, proportionable to the Time proposed. There are several Ways of computing (or answering Questions. about) Simple Interest; as by the single and double Rule of Three (See Page 96, &c.) others make Use of Tables composed at several Rates per Cent. as Sir Samuel Moreland, in his Doctrine of Interest, both limple and compound, all performed by Tables; wherein he hath detected several material Errors committed by Sir Isaac Newton, Mr Kersey upon Wingate, and Mr. Clavil, &c. in the Business of computing Interest, &c. by their Tables, too tedious to be here repeated. But I shall in this Tract take other Methods, and Thew that all Computations relating to Simple Interest are grounded upon Arithmetick Progreffion; and from thence raise such general Theoreins, as will suit with all Cases. In order to that P =any Principal or Sum put to Intereft. Let R=the Ratio of the Rate, per Cont. per Annum. A = the Amount of the Principal, and it's Interest. Note, The Ratio of the Rate, is only the Simple Interest of il. for one Year, at any given Rate; and is thus found. Viz. 100: 6 ::1:0,06 = the Ratio at 6 per Cent, per Annum. Or 100 : 7 :: 1:0,07 = the Ratio at 7 per Cent. &c. Again 100': 7,5:: 1: 0,075 = the Ratio at 7 and 4 per cent. And if the given Time be whole Years; then t= the Number of whole Years: but if the Time be given, be either pure Parts of a Year, or Parts of a Year mixed with Years; those Parts must be turned into Decimals; and then 1 = those Decimals, &c. Now Now the common Parts of a Year may be easily turned or converted into Decimal Parts, if it be considered Day is the sos Part of a Year = 0,00274 ferè That one 3 Month is the part of a Year = 0,0833333 &c. Quarter is the part of a Year = 0,25 Let R= the Interest of il. for one Year, as before. 4 R=the Interest of 1 l. for four Years. And, fo OR for any Number of Years proposed. Hence it is plain, that the Simple Interest of one Pound is a Series of Terms in Arithmetic Progression increasing; whole furt Term and common Difference is Ř, and the Number of all the Terms is t. Therefore the laft Term will always be 1 R = the Interest of il. for any given Term signified by t. As one Pound: is to the Interest of il. :: fo is ang Then { Principal er given Sum : to it's interest. =P Theorem 3 Theorem 4. That is, il:tR::P:RP= the Intereft of P. Then the Principal being added to it's Interest, their Sum will be = 1 the Amount required : which gives this general Theorem. Theorem 1. RP+P=A. From whence the three following Theorems are easily deduced. A ALP Theorem 2. =R. R+I AP RP These four Theorems resolve all Questions about Simple Intereft. Question 1. IVhat will 2561. jos. amount to in 3 Years, one Quarter, 2 Months, and 18 Days, at 6 per Cent. per Annum. Here is given P = 250,5; R = 0,06; and I = 3,46599 For 3 Years = 3 Quære 4. per Theorem I. one Quarter = 0,25 2 Months = 0,16667 = 0,08333 x 2 18 Days = 0,04932 = 0,00274 x 18 Hence t = 3,46599 : x 0,06 = 0,2079594 = IR :** : Then 0,2079594 x 256,5 = 53,341586=+RP And 53,341586 + 256,5 = 309,841586=RP+PEA, That is, 309,841586 = 309 k. 16 si 10 d. being the Answer required... Question Question 2. What Principal or Sum being put to Intereft, will raise a Stock of 309 1. 165. Tod. in three Years, one Quarter, two Months, and i8 Days; at 6 per cent. per Annum ? Or the same Question otherwise stated thus. What is 309 l. 16 s. 10 d. due 3 Years, one Quarter, 2 Months and 18 Days bence, worth in ready Money; abating or discounting 6 per Cent, &c. Here is given A= 309,841586: R=0,06; 1 = 3,46599 (found as before) thence to find P. Per Theorem 2. First 3,46599 x 0,06 = 0,2079594 = + R. Then R+I 1,2079594) 309,841586 = A (256,5 = P; that is, 256,5 = 256 1. 10 s. the Answer required. . Question 3. At what Rate or Interest, per Cent, &c. will 2561. 10 s. amount to 309 l. 165. 10d. in three Years, one Quarter, two Monsbs, and 18 Days ? Here is given, P=256,5; A=309,841586; and t=3,46599 to find Ř. Per Theorem 3. First 309,841586 - 256,5 = 53,341586=A_P. Next 3,46599 x 256,5 = 889,026435 =iR. And + R = 889,026435) 53,341586 (00,06 = the Ratio: 'Then il. : 0,06 :: 120 : 6 = the Rate required. Question 4. In what Time will 2561. 10 s. raise a Stock of (or amount to) 309 l. 16 s. 10 d. at 6 per Cent. &c. Here is given, P = 256,5; A= 309,841586, and R=0,06 to find t. Per Theorem 4. Per Theorem 4. Firit 309,841586 53,341586 = A - P. And 256,5 * 0,06 = 15,39 = R R. Then 15,39) 53,341586 (3,46599 = t; that is 1 = 3 Years and ,46599 Decimal Parts of a Year; which may be brought into common Parts of a Year, thus 0,46599 And 0,08333) 0,21599 (2 Months. 0,25 = one Quarter ,16666 0,21599 0,02074) ,04933 · (18 Days. Hence 153 Years, one Quarter, 2 Months, and 18 Days; the Answer required. It muft needs be easy to conceive, that what is here done at 6 per Cent. may be done at any other Rate of Interest, by forming the Ratio (viz. R) accordingly. SCHOLIU M. |