112 
THE APPLICATION OP ALGEBRA 
PROBLEM LXi 
There are four numbers in arithmetical progression. 
whereof the product of the extremes is 3250 (a J, and 
that of the means 3306 (b J ; what are the numbers ? 
Let the lesser extreme be represented by y s and the 
common difference by л?; then the four required num 
bers will be expressed by у, у f x, у ] 2x, and у f 3x: 
therefore, by the question, we have these two equations* 
viz. 
у x у 4- 3x, or у г -t- зxy — a, and 
у + òc x у 4- 2-т, or у г f Эху f 2Х г = b; w 
the former being taken from the latter, we get 2х г — 
find y from hence, we have given f 1- 3xy = a (by the 
first step); therefore, by completing the square, &c. 
a / 9 A' 2, 3x 
y' v a 4- — — 50: and so the four num 
bers are 50, 55, do, and 65. 
problem lxi: 
The sum (30), and the sum of the squares (30S) of three 
numbers in arithmetical progression being given; to find 
the numbers. 
Let the sum of the numbers be represented by 3b t 
the sum of their squares by c, and the common diffe 
rence by x: then y since the middle term, or number, 
from the nature of the progression, is — b, or -j of the 
whole sum, the least term, it is evident, will be ex 
pressed by b — x t and the greatest by b f x ; and 
therefore, by the question, we have this equation, 
b —V} 1 + b x q- t -f x ~ — c ; which, contracted, 
gives 3— c; whence Qx 1 — c — 3b 1 , and 
Therefore 8, 10, and 12, are 
the three numbers sought,