Question
If a,b,c are in continued proportion, show that: $\frac{a^2+b^2}{b(a+c)}=\frac{b(a+c)}{b^2+c^2}$
✓
Answer
Sine a, b, c are in continued proportion,
$\frac{ a }{ b }=\frac{ b }{ c }$
$\Rightarrow b^2=a c$
Now, $\left(a^2+b^2\right)\left(b^2+c^2\right)=\left(a^2+a c\right)\left(a c+c^2\right)$
$=a(a+c) c(a+c)$
$=a c(a+c)^2 $
$ =b^2(a+c)^2$
$ \Rightarrow\left(a^2+b^2\right)\left(b^2+c^2\right)=[b(a+c)][b(a+c)]$
$\Rightarrow \frac{ a ^2+ b ^2}{ b ( a + c )}=\frac{ b ( a + c )}{ b ^2+ c ^2}$
$\frac{ a }{ b }=\frac{ b }{ c }$
$\Rightarrow b^2=a c$
Now, $\left(a^2+b^2\right)\left(b^2+c^2\right)=\left(a^2+a c\right)\left(a c+c^2\right)$
$=a(a+c) c(a+c)$
$=a c(a+c)^2 $
$ =b^2(a+c)^2$
$ \Rightarrow\left(a^2+b^2\right)\left(b^2+c^2\right)=[b(a+c)][b(a+c)]$
$\Rightarrow \frac{ a ^2+ b ^2}{ b ( a + c )}=\frac{ b ( a + c )}{ b ^2+ c ^2}$
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