If a, b, c are in continued proportion, prove that: a+b b+c =a^2(b-c) b^2(a-b)

a+b b+c =a^2(b-c) b^2(a-b) Since, a, b, c are in continued proportion, a b =b c . Let, a b =b c =k.Then, a = bk and b = ck Hence, a = bk =( ck ) k = ck ^2 & LHS =a+b b+c & LHS =c k^2+c k c k+c & LHS =c k(k+1) c(k+1) & LHS = e k( k+ 1) e( k+ 1) & LHS = k ( I ) & RHS =a^2(b-c) b^2(a-b) & RHS = (c k^2 )^2(c k-c) (c k)^2 (c k^2-c k ) & RHS =c^2 k^4(c k-c) c^2 k^2 (c k^2-c k ) & RHS =c^3 k^4(k-1) c^3 k^3(k-1) & RHS = e^2 k^4( k- 1) e^3( k- 1) & RHS = k ( II ) From (I) and (II), LHS = RHS

If a, b, c are in continued proportion, prove that: a+b b+c =a^2(…

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Question

If $a, b, c$ are in continued proportion, prove that:
$
\frac{a+b}{b+c}=\frac{a^2(b-c)}{b^2(a-b)}
$

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Answer

$
\frac{a+b}{b+c}=\frac{a^2(b-c)}{b^2(a-b)}$
Since, $a, b, c$ are in continued proportion, $\frac{a}{b}=\frac{b}{c}$.
Let, $\frac{a}{b}=\frac{b}{c}=k$.Then, $a = bk$ and $b = ck$
Hence, $a = bk =( ck ) \cdot k = ck ^2$
$
\begin{aligned}
& \text { LHS }=\frac{a+b}{b+c} \\
& LHS =\frac{c k^2+c k}{c k+c} \\
& LHS =\frac{c k(k+1)}{c(k+1)} \\
& LHS =\frac{\not e k(\not k+\not 1)}{\not e(\not k+\not 1)}
\end{aligned}
$
$
\begin{aligned}
& \text { LHS }= k \ldots( I ) \\
& RHS =\frac{a^2(b-c)}{b^2(a-b)} \\
& RHS =\frac{\left(c k^2\right)^2(c k-c)}{(c k)^2\left(c k^2-c k\right)} \\
& \text { RHS }=\frac{c^2 k^4(c k-c)}{c^2 k^2\left(c k^2-c k\right)} \\
& \text { RHS }=\frac{c^3 k^4(k-1)}{c^3 k^3(k-1)} \\
& \text { RHS }=\frac{\not e^2 \not k^4(\not k-\not 1)}{\not e^3(\not k-\not 1)} \\
& \text { RHS }= k \ldots( II )
\end{aligned}
$
From (I) and (II),
$
\text { LHS }=\text { RHS }
$