If a, b, c are in continued proportion then show that a c =a^2+a b+b^2 b^2+b c+c^2

a, b, c are in continued proportion a b =b c Let, a b =b c =k b=c k and a=c k^2 LHS =a c =c k^2 c =k^2 RHS = a^2+a b+b^2 b^2+b c+c^2 = (k^2 c )^2+k^2 c(c k)+(c k)^2 (c k)^2+(c k)(c)+c^2 =k^4 c^2+k^3 c^2+c^2 k^2 c^2 k^2+c^2 k+c^2 =c^2 k^2 (k^2+k+1 ) c^2 (k^2+k+1 ) =k^2 LHS = RHS a c =a^2+a b+b^2 b^2+b c+c^2

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

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Question

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

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Answer

$a, b, c$ are in continued proportion
$\therefore \frac{a}{b}=\frac{b}{c}$
Let, $\frac{a}{b}=\frac{b}{c}=k \quad \therefore b=c k$ and $a=c k^2$
$ \text { LHS }=\frac{a}{c}=\frac{c k^2}{c}=k^2$
$\text { RHS }= \frac{a^2+a b+b^2}{b^2+b c+c^2}$
$=\frac{\left(k^2 c\right)^2+k^2 c(c k)+(c k)^2}{(c k)^2+(c k)(c)+c^2}$
$=\frac{k^4 c^2+k^3 c^2+c^2 k^2}{c^2 k^2+c^2 k+c^2}$
$=\frac{c^2 k^2\left(k^2+k+1\right)}{c^2\left(k^2+k+1\right)}$
$=k^2$
$\therefore \quad \text { LHS }=\text { RHS }$
$\therefore \quad \frac{a}{c}=\frac{a^2+a b+b^2}{b^2+b c+c^2}$