Question

If $a,b,c,d$ are in continued proportion, prove that

$a^{2}c+a{c}^2:b^{2}d+b{d}^2=(a+c{)}^3:(b+d{)}^3$.

Answer 1

Now here $a,b,c,d$ are in continuous proportion, so

$\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{c}=\frac{b}{d}=k$

let it be $k$ where $k$ is a constant.

Given to prove,

$\frac{a^{2}c+a{c}^2}{b^{2}d+b{d}^2}=\frac{(a+c{)}^3}{(b+d{)}^3}$

$\frac{ac(a+c)}{bd(b+d)}=\frac{(a+c{)}^3}{(b+d{)}^3}$

$\frac{ac}{bd}=\frac{(a+c{)}^2}{(b+d{)}^2}$

$\frac{ac}{bd}=\frac{a^2+c^2+2ac}{b^2+d^2+2bd}$

$\frac{b^2+d^2+2bd}{bd}=\frac{a^2+c^2+2ac}{ac}$

$\frac{b}{d}+\frac{d}{b}+2=\frac{a}{c}+\frac{c}{a}+2$

Putting the values we get,

$k+\frac{1}{k}+2=k+\frac{1}{k}+2$

which is true.