Answers

2016-02-28T17:59:56+05:30
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}
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2016-02-28T20:10:31+05:30
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.

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