8. Because the length dispatched (in equal times) is proportional to the Celerities; the Lines of Motion (answering to those equal Times) are to be as 1/_m_, 1/_m_², 1/_m_³, 1/_m_⁴, _&c._ of what they would have been, in the same Times, had there been no Resistance. 9. This therefore is a Geometrical Progression; and (because of _m_ greater than 1) continually decreasing. 10. This decreasing Progression infinitely continued (determining in the same Point of Rest, where the Motion is supposed to expire) is yet of a finite Magnitude; and equal to 1/(_m_ - 1), of what it would have been in so much Time, if there had been no Resistance. As is demonstrated in my Algebra, _Chap._ 95. _Prop._ 8. For (as I have elsewhere demonstrated) the Sum or Aggregate of a Geometrical Progression is (_VR - A_)/(_R_ - 1) (supposing _V_ the greatest Term, _A_ the least, and _R_ the common Multiplier.) That is _VR_/(_R_ - 1) - _A_/(_R_ - 1). Now in the present Case, (supposing the Progression infinitely continued) the least Term _A_, becomes infinitely small, or = 0. And consequently _A_/(_R_ - 1) doth also vanish, and thereby the Aggregate becomes = _VR_/(_R_ - 1). That is (as will appear by dividing _VR_ by _R_ - 1;) _V + V/R + V/RR + V/R³ + &c._ = _VR_/(_R_ - 1);[14] (supposing the Progression to begin at _V_ = 1.) That is (dividing all by _R_, that so the Progression may begin at _V/R_ = 1/_m_:) _V_/(_R_ - 1) = _V/R + V/RR + V/R³ + &c._, That is, in our present Case (because of _V_ = 1, & _R_ = _m_:) 1/_m_ + 1/_mm_ + 1/_m_³ &c. = 1/(_m_ - 1). That is, (putting _n_ = _m_ - 1) 1/_n_ of what it would have been if there had been no Resistance. 11. This infinite Progression is fitly expressed by an Ordinate in the Exterior Hyperbola, parallel to one of the Asymptotes; and the several Members of that, by the several Members of this, cut in continual Proportion. As is there demonstrated at _Prop._ 15. For let _SH_, (_vid._ Fig. 4. Tab. 5.) be an Hyperbola between the Asymptotes _AB_, _AF_: And let the Ordinate _DH_ (in the Exterior Hyperbola, parallel to _AF_,) represent the impressed Force undiminished; or the Line to be described in such time, by a Celerity answerable to such undiminished Force. And let _BS_ (a like Ordinate) be 1/_m_ thereof; which therefore, being less than _DH_ (as being equal to a Part of it) will be farther than it from _AF_. In _AB_ (which I put = 1) let _Bd_ be such a Part thereof, as is _BS_ of _DH_. Now because (as is, well known) all the inscribed Parallelograms, in the Exterior Hyperbola, _AS_, _AH_, _&c._ are equal; and therefore their sides reciprocal: Therefore as _Ad_ = 1 - 1/_m_ (supposing _Bd_ to be taken, from _B_ towards _A_,) to _AB_ = 1, or as _m_ - 1 to _m_: so is _BS_ = (1/_m_)_DH_, to _dh_, which is therefore equal to 1/(_m_ - 1) of _DH_; that is (as will appear by dividing 1, by _m_ - 1,) to 1/_m_ + 1/_mm_ + 1/_m_³, _&c._ of _DH_.[15]