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