Question
If $a, b, c$ are in continued proportion and $a(b -c) = 2b$, prove that: $a-c=\frac{2(a+b)}{a}$
✓
Answer
Since a, b, c are in continued proportion,
$\frac{a}{b}=\frac{b}{c}$
$\Rightarrow b^2 = ac ...(i)$
a(b - c) = 2b
$\Rightarrow \frac{a(b-c)}{b}=2 \ldots$ (ii)
Now,
$\begin{aligned} & \text { R.H.S. }=\frac{2(a+b)}{a} \\ & \Rightarrow \frac{a(b-c)}{b} \times \frac{a+b}{a} \quad \ldots \text { Using equation (ii) } \\ & \Rightarrow \frac{\not a(b-c)}{b} \times \frac{a+b}{\not a} \\ & \Rightarrow \frac{b-c}{b} \times a+b \\ & \Rightarrow \frac{b a+b^2-a c-b c}{b} \\ & \Rightarrow \frac{b a+a c-a c-b c}{b} \quad \ldots .(\text { Using equation (I)) } \\ & \Rightarrow \frac{b a-b c}{b} \\ & \Rightarrow \frac{\not b(a-c)}{\not b} \\ & \Rightarrow \mathrm{a}-\mathrm{c}\end{aligned}$
$\frac{a}{b}=\frac{b}{c}$
$\Rightarrow b^2 = ac ...(i)$
a(b - c) = 2b
$\Rightarrow \frac{a(b-c)}{b}=2 \ldots$ (ii)
Now,
$\begin{aligned} & \text { R.H.S. }=\frac{2(a+b)}{a} \\ & \Rightarrow \frac{a(b-c)}{b} \times \frac{a+b}{a} \quad \ldots \text { Using equation (ii) } \\ & \Rightarrow \frac{\not a(b-c)}{b} \times \frac{a+b}{\not a} \\ & \Rightarrow \frac{b-c}{b} \times a+b \\ & \Rightarrow \frac{b a+b^2-a c-b c}{b} \\ & \Rightarrow \frac{b a+a c-a c-b c}{b} \quad \ldots .(\text { Using equation (I)) } \\ & \Rightarrow \frac{b a-b c}{b} \\ & \Rightarrow \frac{\not b(a-c)}{\not b} \\ & \Rightarrow \mathrm{a}-\mathrm{c}\end{aligned}$
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