If a,b and c are in geometric progression and a+b+c=xb,then prove that x<1 or x>3.

Asked by  | 5th Jul, 2008, 03:41: PM

Expert Answer:

As a,b,c are in geometric progression,

let b=ar and c=ar2

Now ,a(1+r+r2)=xar

Hence x=1+r+1/r

Now by AM >= GM

(r+1/r)/2 >= (r*1/r)1/2 for r>=0

and (r+1/r)/2 <= (r*1/r)1/2 for r>=0
















 

Hence,

      r+1/r >=2 or r+1/r <=-2,Depending on whether r >0 or r<0.

Hence x>3 or x<-1

Answered by  | 18th Jul, 2008, 09:54: AM

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