then, x-y, x, and x+y, will denote the three numbers; .. by the question, 2x2-xy+y2=16, and 2x2-y2 ... by addition, 4x2-xy =17; =33; by transposition, xy=4x2—33; squaring both sides, x2y2-16x-264x2+1089; from the second equation, y=2x2-17; multiplying this equation by x2, x2y2—2x1—17x3 ; .. 16x1-264x2 + 1089=2x1-17x2; by transposition, 14x1-247x2=—1089; completing the square, .. x=3, and y=1; whence, the numbers are 2, 3, and 4. 3. A charitable person relieved four poor women, to the first of whom he gave a certain number of shillings, to the second he gave two shillings more than to the first, to the third he gave two shillings more than to the second, and to the fourth he gave two shillings more than to the third: now the continued product of their money, multiplied in shillings, amounts to £173. 5s. Required the sum given to each. Let x-3, x-1, x+1, and x+3 denote the number of shillings respectively given to each; .. by the qu. x-3xx-1×x+1xx+3=3465, or x-10x+9=3465; completing the square, x*-10x2+25=3481; extracting the root, x2-5=59; .. x=8; whence, he gave 5s. to the first, 7s. to the second, 9s. to the third, and 11s. to the fourth. 4. It is required to find five numbers in arithmetical progression, whose sum is 25 and product 2520. and y the common difference; then, x-2y, x-y, x, x+y, and x+2y will denote the five numbers; .. 5x=25; and x=5. Also, (x2-4y).(x2—y2).x=2520; substituting the value of x in this equation, (25-4y2).(25-y2).5=2520; 125 117 8; comp. the sq. y^2y2+ y extracting the root, y2- 8 .. by taking the negative quantity, y=1; whence, the numbers are 3, 4, 5, 6, and 7. 5. Divide £190 among A, B, and C, so that their respective shares may be in geometrical proportion, and that C's share may exceed A's by £50. Required each person's share. Let x A's share in pounds; then, x+50 = C's share, and 190-2x-50, or 140-2x = B's share ; .. x: 140—2x :: 140–2x:x+50; then 19600-560x+4x2x2+50x; 610x 19600 by transposition, x2 ; 3 185 3 305 extracting the root, x— ± taking the negative value, x=40; whence, the other shares are £60 and £90. 6. Required four numbers in geometrical progression, the sum of the first and second being 20, and the sum of the third and fourth 45. Let x = the first term, and y = the third; = then, 20-x= the 2nd term, and 45—y — the 4th. 9x or 45x-xy=20y-xy, or 20y=45x ; .. y=, and xy=400-40x+x2; substituting the value of y in the second equation, 9x2 Зох =400—40x+x2; .. 3—20—x ; and 3x-40-2x; 9x x=8, and y=18; whence, the numbers are 8, 12, 18, and 27. Or thus: Let x = the first term, and y the common ratio, then, x, xy, xy3, and xy3 will denote the four numbers; .. by the question, x+xy=20, and xy2+xy3=45, or x.(1+y)=20, and xy2.(1+y)=45; whence, the numbers are 8, 12, 18, and 27. 7. Divide £700 among A, B, C, and D, so as their shares may be in geometrical progression, and that the difference between A's and D's may be to the difference between B's and C's as 37 to 12. Required each person's share. Let x = A's share in pounds, and y = the common ratio; then, xy B's, xy C's, and xy3 D's; .. by the question, x+xy+xy2+xy3=700. Also, xy3-x: xy2-xy:: 37:12; .. 12.(xy3—x)=37.(xy2—xy), or 12.(y-1)=37.(y2—y) ; dividing this equation by 12.(y—1), 37 y2+y+1=12¿y ; .. 12y2+12y+12=37y; 25 by transposition, &c. y2 — —y=−1; substituting this value of y in the first eq. x=108; whence, A's share £108, B's £144, C's £192, and D's £256. 8. There is a number consisting of three digits, in geometrical progression, the first of which is to the second as the second to the third, the number itself is to the sum of its digits as 124 to 7, and if 594 be added to it, the digits will be inverted. Required the number. Let x = the first digit, or that in the hundredth's place, and y = the ratio; then x, xy, and ry2 will denote the three digits; .. by the question, 100x+10xy+xy2: x+xy+xy2 :: 124 : 7; dividing the first and second terms by x, 100+10y+y1+y+y2:: 124:7; Theorem 4, 99+9y: 1+y+y: 117: 7; Theorem 10, 11+y: 1+y+y2 :: 137; .. 13y2+13y+13=77+7y; by transposition, 13y2+6y=64; completing the sq. &c. y is found to be equal to 2. Also, by the question, 100x+10xy+xy2+594=100xy2+10xy+x; by transposition, 99x+594=99xy2; dividing by 99, x+6=xy2; substituting the value of y in this equation, x+6=4x; .. x=2; whence, the number is 248. EXAMPLES FOR PRACTICE In Arithmetical and Geometrical Progression. 1. The three sides of a triangle, whose perimeter is 15, are in arithmetical progression, and the sum of the first and second terms, is to the sum of the second and third, as 2 to 3. Required the sides. Ans. 3, 5, and 7. 2. The three sides of a triangle, whose perimeter is 24, are in arithmetical progression, and the product of the first and second, is to the product of the second and third, as 3 to 5. Required the sides of the triangle. Ans. 6, 8, and 10. N |