Question

If a, b, c are in continued proportion , then prove that

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

(ii) $\frac{b}{b+c}=\frac{a-b}{a-c}$

Answer 1

a, b, c are in continued proportion.

$$\begin{gathered} \therefore \frac{a}{b}=\frac{b}{c}=k \\ \Rightarrow a=bk,b=ck \\ \Rightarrow a=bk=ck\times k=c{k}^2 \end{gathered}$$
(i)
$$\begin{gathered} \frac{a}{a+2b}=\frac{c{k}^2}{c{k}^2+2\times ck}=\frac{c{k}^2}{ck\left(k+2\right)}=\frac{{\displaystyle k}}{{\displaystyle k+2}}.....\left(1\right) \\ \frac{a-2b}{a-4c}=\frac{c{k}^2-2\times ck}{c{k}^2-4c}=\frac{ck\left(k-2\right)}{c\left(k^2-4\right)}=\frac{{\displaystyle k\left(k-2\right)}}{{\displaystyle \left(k-2\right)\left(k+2\right)}}=\frac{{\displaystyle k}}{{\displaystyle k+2}}.....\left(2\right) \end{gathered}$$
From (1) and (2), we get

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

(ii)
$$\begin{gathered} \frac{b}{b+c}=\frac{ck}{ck+c}=\frac{ck}{c\left(k+1\right)}=\frac{{\displaystyle k}}{{\displaystyle k+1}}.....\left(1\right) \\ \frac{a-b}{a-c}=\frac{c{k}^2-ck}{c{k}^2-c}=\frac{ck\left(k-1\right)}{c\left(k^2-1\right)}=\frac{{\displaystyle k\left(k-1\right)}}{{\displaystyle \left(k-1\right)\left(k+1\right)}}=\frac{{\displaystyle k}}{{\displaystyle k+1}}.....\left(2\right) \end{gathered}$$
From (1) and (2), we get

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