Question

Find x if x; x+2 and x+b are three consecutive number in GP

Answer 1

Here x , x+2 & x+b are three consecutive numbers in G.P

Therefore

$a_{1} = x ; a_{2} = x+2;a_{3} = x+b$

Therefore $r = \frac{a_{1}}{a_{2}} = \frac{a_{3}}{a_{2}}$

Or, $\frac{x+2}{x} = \frac{x+b}{x+2}$
$(x+2)^2 = x(x+b)$
$x^2 + 2x +4 = x^2 +bx$
$4 = bx -2x$
$4 = x(b-2)$
$x = \frac{4}{b-2}$

Answer 2

according your problem,
the answer is 
if x, x+2,x+b   are in GP then there common ratio is equal then 
a₂/a₁ = a₃/a₂
then x+2/x = x +b/ x+2
= (x+2)² = x² +bx
=x²+4x+4 = x²+ bx
= 4x+4 = bx
= 4x-bx = -4
= x(4-b) = -4
= x = -4 / 4-b

then according to the new text book the answer is 

if the  x , x+2 , x+6 are in GP
then
x+2/x = x+6/x+2
= (x+2)²=x²+6x
= x²+4x+4 = x²+6x
= 4x+4 = 6x
= 4x-6x = -4
= -2x = -4
= x = -4/-2
= x =2
then x,x+2,x+6 are 2,4,8
there are in GP.