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If a, b,and c be three distinct real numbers in gp and a+b+c=xb then b can't be


Zehra Namak 19th May, 2019
Answer (1)
NAVEEN MEENA Student Expert 20th May, 2019

Dear Student, I think you want to find the value for x as a, b and c are given real numbers in gp and as per the given relation, this can be true or false depends on the value of x.

For this relation to hold x should be grater than 3 or less than - 1.

The three numbers are in geometric progression,

(b/a) =r

=> a=b/r and

(c/b) =r

=> c=br
with r= common ratio of the progression.
Since a,b,c are three distinct numbers, r cannot be equal to 1 and - 1

=>a+b+c=b/r + b +br = (b +br +br^2)/r = b(1+r + r^2)/r


If a+b+c=xb then,

xb = b(1+r+r^2)/r

=>. x= (1+r+r^2)/r

If r<0 and r not equal to - 1,

x+1 = (1+r+r^2)/r  +  1 = (1+2r+r^2)/r

=>( (r+1)^2)/r  <0

and x+1<0 => x<-1


If r>0 and r not equal to 1 then,

x-3 = (1+r+r^r)/r  - 3 =(1-2r+r^2)/r

=>. ((r-1)^2)/r    >0

and x-3>0   =>  x>3.

Hope this helps :)

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