Question

if a, b, c are in gp then find the value of b-a/b-c + b+a/b+c​

Answer 1

Let a=1 b=2 c=4

By taking 1,2,4,.... As our gp

b-a/b-c + b+a/b+c

=(1/-2)+(3/6)

=-1/2 + 1/2

=0

Answer 2

Given:-

As , it is given that

a,b,c -----> in G.P

Solution:-

∴$\frac{b}{a} = \frac{c}{b}$

= $b^{2} = ac$

We have ,

$\frac{b-a}{b-c} + \frac{b+a}{b+c} = ?$

divide by a

we get,

= $\frac{\frac{b}{a} -1 }{\frac{b}{a} - \frac{c}{a}} + \frac{\frac{b}{a} +\frac{a}{a} }{\frac{b}{a} + \frac{c}{a} }$

= $\frac{1}{c} \frac{\frac{b-a}{a} }{\frac{1}{b} - \frac{1}{a} } + \frac{1}{6} \frac{b+a/a}{\frac{1}{b} - \frac{1}{a} }$

= $\frac{1}{6} [ \frac{-ab}{a} + \frac{ab}{a} ]$

= $\frac{1}{c} *0$

= $0$

hence the value of your Question is = 0

thank you