a, b, c are in G.P.

So, b = ar

and c = ar²

given   a + b + c = xb

$$ \Rightarrow $$  a + br + ar² = x(ar)

$$ \Rightarrow $$  1 + r + r² = xr

$$ \Rightarrow $$  x = 1 + r + $${1 \over r}$$

let sum of r + $${1 \over r}$$ = M

$$ \therefore $$  r² + 1 = Mr

$$ \Rightarrow $$  r² $$-$$ Mr + 1 = 0

this quadratic equation will have

real solution when discriminant is $$ \ge $$ 0

$$ \therefore $$  b² $$-$$ 4ac $$ \ge $$ 0

M² $$-$$ 4.1.1 $$ \ge $$ 0

$$ \Rightarrow $$  M² $$ \ge $$ 4

M $$ \ge $$ 2 or M $$ \le $$ $$-$$ 2

$$ \therefore $$  M $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 2] $$ \cup $$ [2, $$ \propto $$)

As   x = 1 + r + $${1 \over r}$$

= 1 + M

$$ \therefore $$  x $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 1] $$ \cup $$ [3, $$ \propto $$)

$$ \therefore $$  x can't be 0, 1, 2.