I Add equations 1, 2, and 3 together, and the sum will be + ̧; from eq. 4 subtract eq. 1, 2, and 3 severally, and the re 2 r EXAMPLES.-1. A and B can unload a waggon in 3 hours, B and C in 24 hours, and A and C in 24 hours; how long will each be in doing the same by himself? 2. A quantity of provisions will serve A and B 8 months, A and C 9 months, and B and C 10 months; how long would the same quantity serve each person singly? Ans. A 14 m. 2010 days, B 17 m. 1634 days, C 23 m. 6÷1 days, reckoning 30 days to a month. 45. It is required to divide the number 22 into three such parts, that once the first, twice the second, and thrice the third being added together, the sum will be 47, and the sum of the squares of the parts 166? Let x, y, and z, denote the three parts respectively, a=22, b =47,c=166; then by the problem x+y+z=a, x+2y+3z=b, and x2+ y2+z2=c; subtract the first from the second, and y+2z =b-a, whence y=b-a-2 z; subtract double the first from the second, and z-x=b-2a, whence x=z+2a−b; let f=b-a, gb-2a; these values being substituted in the two latter equations, they become y=f-2z, and x=z-g; substitute these values for y and x in the third given equation, and it will become 22-2gz 2f+g c-f2-g2 +g2+ƒ2¬4 ƒz+4z2+z2=c, or z2_2.ƒ+82 3 C 6 2f+g, and the latter equation becomes z2—hz= 3 -; put h= 2 which by completing the square, &c. it becomes z: ± c-f2—g2 6 h2 4 (which, by restoring the values of c, f, g, and h, viz. c=166, f=b-a=47—22=25, g=b−2 a=47—44= =9, whence x=(z-g=) 6, and y= (f-2 z=) 7. 46. Required the values of x and y in the following equations, viz. x+x3y+x2y2+xy3+y^=211=a; and x2+x®y2+x+y*+ xy+y=11605=b? Divide the second by the first, add the quotient to, and subtract it from the first, and the results will be (2x2+2x2y2+2y*= 1 b = a 2a' then will the two equations, above derived, become {x* +x2y2+y*=x2+yo]2 —x2y2=) s°—p2=m, and (x2+yo .xy=) n _sp=n, ··· p=~ ; this being squared, and the square added to so — p2=m, gives s2=m++, or st—ms2=n2, ··· s= √↓· n =13, and p=(—=)6. Now since (s=) x2+y2=13, and (p=) xy=6, if the square root of the sum and difference, of the former and double the latter be taken, we shall thence obtain x=3, and y=2. 47. Given the sums, and the product=p, of any two numbers, to find the sum of their nth powers? Let x and y represent the two numbers, then will x+y=s, and ry=p. First, (x+y)2=) x2+2xy+y2=s2, and 2xy=2p; subtract the latter from the former, and x2+y=s-2p=the sum of the squares. Secondly, x2+y2.x+y=s2—2p.s, or x3 +xy.x+y +y3=s—2 sp, which (by substituting sp for its equal_xy.x+y) becomes x3+sp+y3=s3 —2 sp, ··· x3+y3=s3-3 sp=the sum of the cubes. Thirdly, x3+y3.x+y=s3—3 sp.s, or x1+xy.x2 + y2+ y=st-3 s2p, which (by_substituting p.s2-2p for its equal y.x+y) becomes x++p.s2-2p+y+=st-3 s2p, . x++y+= (s— 3sp-p.s2-2p=) st—4 s3p+2 p2 the sum of the biquadrates. In like manner it may be shewn, that s5-5 s3p+5 sp2=the sum of the fifth powers; s6—6 s1p+9 s'p2 -2 p3=the sum of the sixth powers, &c. = By comparing together these several results, the law of continuation will be manifest; for it appears from the foregoing process, that The sum of any powers is found by multiplying the sum of the next preceding powers by s, and from this product subtracting the sum of the powers next preceding those multiplied by p. Thus, the sum of the 4th powers sx sum of the cubes-px sum of the squares. The sum of the 5th powers=s× sum of the 4th powers—px sum of the cubes. The sum of the 6th powers=s× sum of the 5th powers—px sum of the 4th powers, &c. &c. Hence the sum of the nth powers of x and y will be as follows; n-2 n -- x+y"s"-n.s p+n. -.S p2. -n. 3 n-4 2 2 3 4 Then will a+a+d+a+2d+a+3d+, &c. to a+n-1.d be an increasing series of terms in arithmetical progression. And z+z-d+z−2d+z−3d+, &c. to z-n-1.d will be a decreasing series in arithmetical progression. 14. Now since in the increasing series a+n-1.d=the greatest term, and z=the greatest term by the notation, therefore z=a+ n-l.d (THEOREM 1.) Whence by transposition, &c. a=z—n—1.d (THEOR. 2.) d= (THEOR. 3.) and n== +1 (THEOR. 4.) n- -1 d Whence, of the first term, last term, number of terms, and difference, any three being given, the fourth may be found by one of these four theorems. 15. Next, in order to find s, and to introduce it into the foregoing theorems, let either of the above series, and the same series inverted be added together; and since the sum of each series is= s by the above notation, the sum of both added together, will evidently be 2 s. Thus, The series... a+a+d+a+2d+a+3d+&c.=s. The series inverted a+3d+a+2d+a+d+a... =s. Their sum ..... Qa+3d+2a+3d+2a+3d+2a+3d=2s That is (2a+3d.n, or a+a+3 d.n, or, since a+3 d= 2) a+z.n n a+z.n=2s, whence s=(- =) a+z. — (THEOR. 5.) From this 28 2 2 equation are derived a=-z (THEOR. 6.) z=- -a (THEOR. 7.) n n (THEOR. 8.) Also by equating the values of z in 28 theorems 1 and 7, (viz. a+n—1.d=——a,) we obtain a= n 28 2 a (THEOR. 10.) n.n-1 n-1 n n-] s=n.2a+n−1.d (THEOR. 11.) and n= i+n−1. (THEOR. 12.) ÷d—a+ √÷d—a'2+2ds d 16. In like manner, by equating the values of a in theorems 2 s 2 and 6, (viz. z—n—1.d=23—z,) we derive z= S n-1 + .d 2 equating the values of n in theorems 4 and 8, we have whence z=√a—d2+2 ds—÷d (THEOR. 17.) a= (THEOR. 16.) and d 17. Hence any three of the five quantities a, z, d, n, s, being given, the other two may be found; also if the first term a=0, any theorem containing it may be expressed in a simpler manner. 18. The following is a synopsis of the whole doctrine of arithmetical progression, wherein all the theorems above derived are brought into one view. |