Finally: something new!

[A$$Bandit]

I got this for h/w today:

If a, b and c are non-zero real numbers, and -
a, b, c, are in arithmetic progression
a, c, b, are in geometric progression
Find a numerical value for the common ratio of the geometric progression.

I keep getting two equations with 3 unknowns in...

[Kalkewl8ter]

I don't know if this is any help at all, but I can prove that b>=c by the following:

Let d be the common difference in arithmetic sequence a, b, c. Let r be the common ratio in geometric sequence a, c, b.
b=a+d=a*(r^2)
c=a+2*d=a*r
By AM>=GM,
(a+b+c)/3>=(a*b*c)^(1/3)
Substituting values for b and c:
(a+a+d+a+2*d)/3>=(a*a*(r^2)*a*r)^(1/3)
a+d>=ar
Thus,
b>=c

[Kalkewl8ter]

This also means that d<=0, and a>=b.
Since a>=b>=c, and a, c, b are in geometric progression,
(-1)<=r<0. The logic behind this is that, since from a to c, the terms declined, and from c to b, the terms grew, so r must be non-positive. Since a, b, and c are non-zero, r cannot be 0, so r is now negative. And if r<-1, b>a, which is not true, so, by contradiction, r>=-1.

Since r cannot be 1, a<>b<>c, thus, a>b>c and d<0.

We now use the following equations:
a*(r^2)=a+d (1)
a*r=a+2*d (2)
(2)-(1):
d=a*r-a*(r^2)
d=a*r*(1-r)
We know that d<0, therefore,
a*r*(1-r)<0
r<0, and 1-r>0, therefore,
a>0, b>0, and c<0.