Question
If $a, b, c$ and dare in continued proportion, then prove that
ad $(c^2 + d^2) = c^3 (b + d)$

Answer

$a d\left(c^2+d^2\right)=c^3(b+d) $
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k $
$\Rightarrow c=k d $
$b=k c=k^2 d $
$a=k b=k^3 d $
$a c\left(c^2+d^2\right)=c^3(b+d)$
LHS
$a c\left(c^2+d^2\right)$
$=k^3 d x c\left(k^2 d^2+d^2\right)$
$=k^3 d^3\left(k^2 d+d\right)$
RHS
$c^3(b+d) $
$=k^3 d^3\left(k^2 d+d\right)$
LHS $=$ RHS. Hence, proved.

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