Question
If a, b, c, d are in continued proportion, prove that:
$\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}$

Answer

Since $a, b, c, d$ are in continued proportion then
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k $
$\Rightarrow a = bk ^2 b = ck , c = dk $
$\Rightarrow a = ck ^2 $
$\Rightarrow a = dk ^3, b = dk ^2 \text { and } c = dk $
$\text { L.H.S. } $
$=\sqrt{a b}-\sqrt{b c}+\sqrt{c d} $
$=\sqrt{d k^3 \cdot d k^2}-\sqrt{d k^2 \cdot d k}+\sqrt{d k \cdot d} $
$=d \cdot k^2 \sqrt{k}-d k \sqrt{k}+d \sqrt{k} $
$=\left(k^2-k+1\right) d \sqrt{k} .$
$\text { R.H.S. } $
$=\sqrt{(a-b+c)(b-c+d)} $
$=\sqrt{\left(d k^3-d k^2+d k\right)\left(d k^2-d k+d\right)} $
$=\sqrt{d \times d \times k\left(k^2-k+1\right)\left(k^2-k+1\right)} $
$=\left(k^2-k+1\right) d \sqrt{k} $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.

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