TO THE RESOLUTION OF PROBLEMS. 
117 
Let the common difference be denoted by 2a, and 
the lesser extreme by y — 3a; then, it is plain, -the 
other three terms of the progression will be expressed 
by y — x, y f x, and y f 3a’ respectively ; and so, by 
the question, we have 
y — 3a|' 4 y — x\ L 4- y + a|* 4 y 4 3a|* — a, and 
y — 3a x y — a x y 4- a x y 4 3a — b, 
tliat is, by reduction, 
4у г 4 20a 2 = a, and 
у 4 — Юу 2 а г 4- да 4 — b; 
from the former of which if — — 5a*: and there 
fore y* — f T a'- — iax z + 25a 4 : these values being 
substituted in the latter, we have T Va 2 — £ах г + 25a 4 
5ax^ 
— ¿ax 2 + 50a 4 4- 9a 4 — b, and therefore a 4 — 
84 
_0_ 
84 
Гб x 84’ W ^ ence ’ completing the square 
~ -4 ) 
4 X 84 X 84 1 84 84 . X 84 
4 X 84 X 84 
84Ô 4 a*. 
84 X 84’ 
therefore a 2 
— 3a 2 ) is also known. 
PROBLEM LXVI 
The difference o f the means (a), and the difference of 
the extremes fbj of four numbers in continued geometri 
cal proportion being given; to find the numbers. 
Let the sum of the means be denoted by a; then the 
greater of them will be denoted by X , and the les- 
ser by whence, by the nature of proportionals, it 
1 3