Question

algebra question
please help me as well as ​

Answer 1

a, b, c are in continued proportions that means

$\frac{a}{b} = \frac{b}{c} =\frac{c}{a}$

This implies that:

b^2 =ac

c^2 =ab

a^2 =bc

Given equation:

$\frac{a}{a+2b}=\frac{a-2b}{a-4c}$

Consider LHS of this equation, we get

$\frac{a}{a+2b} = \frac{a}{a + 2b} \times \frac{a - 2b}{a - 2b} \\ = \frac{a(a - 2b)}{a {}^{2} - 4b {}^{2} }$

$Put b^{2} =ac$

$\frac{a(a-2b)}{(a^{2}-4ac)} = \frac{a - 2b}{a - 4c}$

Which is equal to RHS

Therefore,

LHS = RHS

Hence, proved.