Question

a,b,c are three distinct real numbers,which are in g.p and a+b+c=xb then
a)x<-1 (B) -1<x<2 (c) 2<x<3 (d)x>3

Answer 1

Answer:

If multiple options are correct then (a) and (d)

If only one option is correct then (b)

Step-by-step explanation:

Since a, b, c are in GP

Therefore

b=ar and c=ar², where r is the common ratio

Given

$a+ b+c=xb$

$\implies a+ar+ar^2=arx$

$\implies 1+r+r^2=rx$

$\implies (1+r+r^2)/r=x$

Since a, b, c are distinct

Therefore r ≠ 1

Case (1) When r < 0

$x=\frac{1+r+r^2}{r}$

or, $x+1=\frac{1+r+r^2}{r}+1$

or, $x+1=\frac{1+r+r^2+r}{r}$

$\implies x+1=\frac{(r+1)^2}{r} < 0$   (∵ (r+1)² is positive and r is negative)

$\implies x+1 < 0$

$\implies x < -1$

Case (2) when r > 0

$x-3=\frac{1+r+r^2}{r}-3$

or, $x-3=\frac{1+r+r^2-3r}{r}$

\implies $x-3=\frac{1-2r+r^2}{r}$

$\implies x-3=\frac{(1-r)^2}{r} > 0$

$\implies x-3> 0$

$\implies x> 3$

Thus

either x < -1 or x > 3

Answer 2

Answer:

option (a)  and (d) both are correct

Step-by-step explanation:

A,b,c are three distinct real numbers,which are in g.p and a+b+c=xb then

a)x<-1 (B) -1<x<2 (c) 2<x<3 (d)x>3

Let r be the common ratio

since a,b,c are in G.P, we can write them as $\frac{b}{r}\,\:b,\:br$

Now,

$a+b+c=xb$

$\implies\:\frac{b}{r}+b+br=xb$

$\implies\:b(\frac{1}{r}+1+r)=xb$

$\implies\:\frac{1+r+r^2}{r}=x$

$\implies\:1+r+r^2=rx$

$\implies\:r^2+(1-x)r+1=0$......(1)

since a,b,c are real, r is real number

$\implies\:$The discriminant of the equaion (1) $\bf{b^2-4ac\:>\:0}$

$(1-x)^2-4(1)(1)\:>\:0$

$1+x^2-2x-4\:>\:0$

$x^2-2x-3\:>\:0$

$(x-3)(x+1)\:>\:0$

$\implies\:\boxed{x>3\:or\:x<-1}$