If a, b, c are in continued proportion then show that, (a+b)^2 a …

Question

If $a, b, c$ are in continued proportion then show that, $\frac{(a+b)^2}{a b}=\frac{(b+c)^2}{b c}$.

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Answer

$a, b, c$ are in continued proportion. Let $\frac{a}{b}=\frac{b}{c}=k$.
$\therefore b=c k, \quad a=b k=c k \times k=c k^2$
Substituting values of $a$ and $b$.
$ \text { LHS }=\frac{(a+b)^2}{a b}=\frac{\left(c k^2+c k\right)^2}{\left(c k^2\right)(c k)}=\frac{c^2 k^2(k+1)^2}{c^2 k^3}=\frac{(k+1)^2}{k}$
$\text { RHS }=\frac{(b+c)^2}{b c}=\frac{(c k+c)^2}{(c k) c}=\frac{c^2(k+1)^2}{c^2 k}=\frac{(k+1)^2}{k}$
$\therefore \text { LHS }=\text { RHS. } \quad \therefore \frac{(a+b)^2}{a b}=\frac{(b+c)^2}{b c}$

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