Let x/y=k, and y=kx.
(A) If a,b,c form an arithmetic sequence then a+c=2b. Then
x/y+xy=2(x+y) => k+ky^2=2y(k+1)
Solve for k:
k=2y/(y1)^2=2/(y1)*y/(y1)
k is integer only if y=2, then k=4, x=8
Arithmetic sequence: 4, 10, 16
(B) If a,b,c form a geometric sequence then ac=b^2. Then
x/y*(x+y)=(xy)^2 => ky(k+1)=y^2(k1)^2 =>
k(k+1)=y(k1)^2
Solve for y
y=k(k+1)/(k1)^2
y is integer only if
1) k=2, then y=6 and x=12
Geometric sequence: 2, 6, 18
2) For k>2, k and k1 are relatively prime, so (k+1)/(k1)^2
has to be integer, only solution is k=3, and y=3, x=9
Geometric sequence: 3, 6, 12

Posted by Art M
on 20141010 22:17:10 