to find the Values of all the unknown Quantities, because they afford fuch Variety, as being well oblerved by a Learner, will be found very useful in the Solution of moft Queftions. Note, I have chofe to use the fame Numbers for the refpective Value of each Quantity throughout all the Questions, because they will be more fatisfactory in proving the Work than various Numbers would have been. Not but that any Numbers may be taken at Pleafure, provided that the Number reprefented by a, be greater than that by e, &c. I have omitted the Numerical Calculations purely for the Learner to practise on. Question 16. There are two Numbers, the Sum of their Squares is 2 3 6 8; and the greater of them is in Proportion to the lefs, What are these Numbers? as 6 to 1. Question 17. There are three Numbers in continued Proportion, the Sum of the Extreams is 156, and the Mean is 72; What are the two Extreams? That is, Suppofe a. m. e in, and m=72. Question 18. There are three Numbers in, their Sum is 74, and the Sum of their Squares is 1924; What are those Numbers? That is, a, e, y are in itate+y==74 zaa+ee+yy=z=1924 Quære a, e, y. ·3ae::e: y 4ay=ee Then 5. 2ee 4 6+7 8 aa+zay+yy=z + ee 522 9aa+zay+yy=ss-2 setee 8 and 9 10%+ee=ss=2se fee 10 +11 2 se=ss-Z Note, In all Questions about continual Proportionals, (either Arithmetical or Geometrical) where three Terms are fought, the Mean is the cafieft found firft (as above) and if all the Terms be Affirmative, then it is equal whether the first or laft Term be the greatest. Question Question 19. There are three Numbers in their Sum is 76; and if the Sum of the Extreams be multiplied into the Mean, that Product will be 1248; What are thofe Numbers? { 24 52. per Theorem 3. Chap. 8. Or a=16 6} { and y = 36 17 | 18y = 52 36 = 16 ) { and y = 36 N. B. If you take es+ 4ss-p=52 (at the 10th Step) then it will be 765224a+y, which is impoffible, viz. that the Mean fhould be greater than the Sum of the two Extreams. Therefore it must be es√ss—p=24. (See page 201.) Question 20. There are three Numbers in Arithmetical Progreffion, the first being added to twice the fecond, and three times the third, their Sum will be 62; and the Sum of all their Squares is 275; What are thofe Numbers? Suppofe And { a, e, in Arithmetical Progreffion. 3aatee+3y=275 Then 4a+y=2e, per Sect. 1. Chap. 6.. 2 4 5÷2 256 + 52e+27=62-2e 7y=31-2 e 82 | 9|aα=16ee248e+961 1647 -1647 372e= -4 242 17+22186= = §2 + ;= 9, or 81⁄2 the Mean. 8 and 19 20 36—31=5, or 34 — 31 = 3 $ 7 and 21 22y = 31 — 18 = 13, or 31 — 173=13$ Question 21. There are three Numbers in Arithmetical Progreffion; the Square of the first Term being added to the Product of the other two is 576; the Square of the Mean being added to the Product of the two Extreams, make 612; and the Square of the laft Term being added to the Product of the firft into the fecond, is 792: What are thofe Numbers? Suppofe 11a, e, y in Arith. Progref. as before. Then I '.' 5x e 3 e e+ya=612 by the Queftion. 5a+y=2e, per Sect. 1. Chap. 6. 2 +4 7aa+ye+yy+a=1368 768aa+yy=1368-2ee 3 -ee 9 x 2 10 2 y a 1224 · -2.ee 8 +10 11 aa+zy a + y y = 2592 — 4e e Question 22. It is required to find two fuch Numbers, that the Sum of their Squares may be 8226; and their Product being added to the Square of the leffer, may be 69211. Viz. {| 1 | aa+ee = 82261} Quære a and 1-2 e S 8xaa 942610 aa+1703025=8226,5 a a—a* 9+a2 10 2a4—2610aa17030258226,5 aa 10+ 11 204 10836,5 aa =— 1703025 11÷212 at — 5418,25 a a=-851512,5 (5,765 12 C134-5418,25 aa+7339358,26562=648784 14+27 &c. 15 aa 2709,125+2547,125 7,125) 13 m2 14 aα-2709,1256487845,765625-254 Suppofe 16 aa2709, 125 +2547,125 =5256,25 Then 174✓ 5256,25=72,5 Or let 19aa2709,125-2547,125=162 Then 21e= Therefore 162-1305 which is impoffible. 12,72 a = 72,52 as at the 17th and 18th Steps. And le = 54,55 This Queftion may be performed with lefs Trouble, by fubftituting Letters for the known Numbers. Viz. {aa+e=x? Then let z-da amae, 8cc.. aa+11=2 Ff Question |