SOLUTION: if a,b,c are three distinct real numbers in geometric progression and a+b+c=xb then prove that x is greater than3 or less than -1

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Question 775436: if a,b,c are three distinct real numbers in geometric progression and a+b+c=xb then prove that x is greater than3 or less than -1
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The three numbers are in geometric progression, sp
b%2Fa=r<-->a=b%2Fr and
c%2Fb=r<-->c=br
with r= common ratio of the progression.
Since a,b,c are three distinct numbers,
r%3C%3E1 and r%3C%3E-1

a%2Bb%2Bc=b%2Fr%2Bb%2Bbr=%28b%2Bbr%2Bbr%5E2%29%2Fr=b%281%2Br%2Br%5E2%29%2Fr
If a%2Bb%2Bc=xb xb=b%281%2Br%2Br%5E2%29%2Fr --> x=%281%2Br%2Br%5E2%29%2Fr

If r%3C0 and r%3C%3E-1,
x%2B1=%281%2Br%2Br%5E2%29%2Fr%2B1=%281%2B2r%2Br%5E2%29%2Fr=%28r%2B1%29%5E2%2Fr%3C0
and x%2B1%3C0 --> highlight%28x%3C-1%29

If r%3E0 and r%3C%3E1,
x-3=%281%2Br%2Br%5E2%29%2Fr-3=%281-2r%2Br%5E2%29%2Fr=%28r-1%29%5E2%2Fr%3E0
and x-3%3E0 --> highlight%28x%3E3%29