TO THE RESOLUTION ÔF PROBLEMS» 
109 
Will be x -f l: moreover, the whole price of the pep« 
per will be crowns, and that of the ginger 
therefore, by the question, 
120 
x 
120 
x -f 1 
rr 6; whence 
120.r 120— 120a: — 6r* + 6a?, and therefore x z + x 
— 20; which, solved, gives x = 4 =r the pounds of 
pepper, and x -f l — 5 — those of ginger» 
PROBLEM LV. 
To find three numbers in arithmetical progression„ 
whereof the sum of the squares shall be 1232 (aJ, and 
the square of the mean greater than the product of the 
two extremes by 16 (bj. 
Let the mean be denoted by x, and the common dif 
ference by y; then the numbers themselves will bear—'y t 
x 4 and x -4- y; and so, by the problem, we shall have 
these two equations, 
x —yf + x 1 + x + y\~ zz a, and 
x* — x — y x x + y + b: these, contracted, become 
3x z + 2if zz «, and x z zz x z — y z + b; from the latter 
whereof we get y 2 — b zz 16; and consequently y zz 
\/b — 4; which, substituted for y in the former, gives 
3x* -f 2b zz a; whence x 1 
and therefore 
20; so that the three required 
numbers are 16, 20, and 24. 
v s 16* + 20 x + 24 7 rr 1232, 
1 ° r j 20 7 — 16 X 24 ~ 16. 
PROBLEM LVI* 
To find two numbers whose difference shall be 10 f aj t 
and if600 fbj be divided by each of them, the difference 
of the quotients shall also be equal to 10 f a J> 
The lesser number being represented by x, the greater 
will be represented by x -f a; and therefore, by the prob 
lem ~ —7— = a; which, freed from fractions. 
x x -f a