Or if any fingle Power of the unknown Quantity (a) is found equal to thofe that are known, then the respective Root of the known Quantities is the Anfwer. As in the firft Four Examples of Sect. 6. But when the Powers of the unknown Quantity are either mix'd with their Root, as aa + badd, &c. Or do confift of different Powers; as a da cc, &c. Then they are call'd adfected Equations; the manner of refolving which, that is, of finding the Value of a, has been fhewed in Part IV. Chap. II. And other Methods for the fame purpose shall be hewed further on. Of PART vui. Proportional Quantities both Arithmetical, Geometrical, and Mulical. WE CHAP. I. Of Arithmetical Propoztion. Definition. Hen any Rank or Series of Numbers or Quantities, do either increase or decrease, by an equal Interval or com mon Difference or Excess, they are faid to be in Arithmetical Progreffion continued. In the former there is a continual Increase; in the latter a continual Decreate by 2, which is the common Difference or Excels. And univerfally putting a for the 1st Term, and e for the common Excess or Difference, the Terms will be a, ate, a + ze, a + ze, &c. Increasing. a, a—e, 4 — 2e, a 3e, &c. Decreasing. a But the most Simple and Natural Progreffion is that which be gins with e, as e 20, 36, 4e, se, &c. o, —e, —2o, —3e, —4o, —5e, &c. When it begins with any other Term (as a in the former Progreffions) it is really a Compound of two Progreffions, one of Equals (a, a, a, a, &c.) and the other of Arithmetical Proportionals (o, e, 2c, 3e, 4e, &c.) But But when the firft Term exceeds, or is exceeded by, the 2d Term by the fame Number or Quantity, that the 3d exceeds, or is exceeded by, the 4th Term; but not by the fame that the 2d exceeds, or is exceeded by, the 3d Term; then that is faid to be a Difcontinued or Disjun& Arithmetical Progreffion. So 1, 3, 9, 1'1 la, a + e, a +7e, age are faid to be Disjunct or Discontinued Arithmetical Proportions. Now in order to the finding out how to refolve Questions concerning thefe Proportionals. the leaft Term. the greatest Term. Lete the common Excels. The Sum of the Extreams (i. e. of the leaft and greateft Terms) is equal to the Sum of any two Means that are equally diftant from thofe Extreams (i. e. equal to the Sum of the lefs but one, and greateft but one, or lets but two, and greatest but two, or lefs but three, and greatest but three, &c. Terms) of any Arithmetical Progreffion, either continued or discontinued; and confequently, if the Number of Terms be odd, the double of the middle Term is equal to the Sum of the two Extreams, or if the Number of Terms be even, the Sum of the two Extreams is equal to the Sum of the two middle Terms. }=~ Greatest Term y a Leaft Term but one 4+ c} = a + x Greatest Term but one y Least Term but two a +203 Greatest Term but two y --- Leaft Term but three a +30=4+y. N. B. What follows relates to &c. QE. D. Scholium Scholium.1. Wherefore by proceeding with the firft Series of thefe Terms, towit a, ae, atze, a + 3e, &c. to the n Term, which is y; alfo with the other Series of Terms, towityy — e, y —ze, y3e, &c. to the n Term likewife, which is 4, and then Adding both Series into one Sum, the faid Sum will be equal to double the Sum of either Series, and alfo to n times a+}, that is to fay And the Value of s is found in the next Line before the preceeding Corollaries, Scholium 2. and Theorem 2. In the foregoing Decreafing ga, a +e, a + 2e, a+ 3e, &c. Series, viz. it is easy to perceive that the common Difference e is lo often Subftracted in the laft Term of the Series, as are the Number of Terms except the firft; that is the first Term < Added hath no difference Subftracted in it; but the last hath so many times e Added ted} {substracted Subftracted S } in it, as are the Number of Terms from the firft to the laft Inclufive, confequently the difference of the Extreams is the Number of all the Terms less Unity or 1; that is Then na +nne ne 25 na by Multiplication, And 2sana + nne 2. Therefore sna+ ne by Tranfpofition. Given a, e,y; required n and s. Solution. te I. n= Then 25 a by Theorems 2d and ist. 2es=yyaa+ea+ey by Multiplication. 2. Therefore s yy aa + ea + ey Therefore nne 2nane 2s by Multiplication and Tranf pofition. |
