# Prove that

Question:

If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$ then show that a, b, c, d are in GP.

Solution:

$\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$ (Given data in the question) $\rightarrow$ (1)

Cross multiplying (1) and expanding

$(a+b x)(b-c x)=(b+c x)(a-b x)$

$a b-a c x+b^{2} x-b c x^{2}=b a-b^{2} x+a c x-b c x^{2}$

$2 b^{2} x=2 a c x$

$b^{2}=a c \rightarrow(i)$

If three terms are in GP, then the middle term is the Geometric Mean of first term and last term.

$\rightarrow b^{2}=a c$

So, from (i) b, is the geometric mean of a and b

So, a, b, c are in GP.

$\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}-\mathrm{dx}}$ (Given data in the question) $\rightarrow(2)$

Cross multiplying (2) and expanding,

$(b+c x)(c-d x)=(c+d x)(b-c x)$

$b c-b d x+c^{2} x-c d x^{2}=c b-c^{2} x+b d x-d c x^{2}$

$2 c^{2} x=2 b d x$

$c^{2}=b d \rightarrow($ ii $)$

So, from (ii), c is the geometric mean of b and d.

So, $b, c, d$ is in GP.

$\therefore \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in GP.