If a, b, c, d are in continued proportion, prove that: (a-b c +a-c b )^2- (d-b c +d-c b )^2=(a-d)^2 (1 c^2 -1 b^2 )

a, b, c, d are in continued proportion & a b =b c =c d = k (say) & c = dk , b ^ ck = dk k = dk ^2, & a = bk = dk ^2 k = dk ^3 & L.H.S. = (a-b c +a-c b )^2- (d-b c +d-c b )^2 & = (d k^3-d k^2 dk +d k^3-d k d k^2 )^2- (d-d k^2 dk +d-d k d k^2 )^2 & = (d k^2(k-1) dk +d k (k^2-1 ) d k^2 )^2- (d (1-k^2 ) dk +d (1-k^2 ) d k^2 . & = (k(k-1)+k^2-1 k )^2- (1-k^2 k +1-k k^2 )^2 & = (k^2(k-1)+ (k^2-1 ) k )^2- (k (1-k^2 )+1-k k^2 )^2 & = (k^3-1 )^2 k^2 - (-k^3+1 )^2 k^4 & = (k^3-1 )^2 k^2 - (1-k^3 )^2 k^4 & = (k^3-1 k^2 )^2 (1-1 k^2 ) & = (k^3-1 )^2 (k^2-1 ) k^4 & = (k^3-1 )^2 (k^2-1 ) k^4 & R.H.S. =(a-d)^2 (1 c^2 -1 b^2 ) & = (d k^3-d )^2 (1 d^2 k^2 -1 d^2 k^4 ) & =d^2 (k^3-1 )^2 (k^2-1 d^2 k^4 ) & = (k^3-1 )^2 (k^2-1 ) k^4 & L.H.S. = R.H.S.

If a, b, c, d are in continued proportion, prove that: (a-b c +a-…

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Question

If $a, b, c, d$ are in continued proportion, prove that:
$
\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^2-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^2=(a-d)^2\left(\frac{1}{c^2}-\frac{1}{b^2}\right)
$

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Answer

$a, b, c, d$ are in continued proportion
$
\begin{aligned}
& \therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k \text { (say) } \\
& \therefore c = dk , b ^{ ck }= dk \cdot k = dk ^2, \\
& a = bk = dk ^2 \cdot k = dk ^3 \\
& \text { L.H.S. }=\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^2-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^2 \\
& =\left(\frac{d k^3-d k^2}{ dk }+\frac{d k^3-d k}{d k^2}\right)^2-\left(\frac{d-d k^2}{ dk }+\frac{d-d k}{d k^2}\right)^2 \\
& =\left(\frac{d k^2(k-1)}{ dk }+\frac{d k\left(k^2-1\right)}{d k^2}\right)^2-\left(\frac{d\left(1-k^2\right)}{ dk }+\frac{d\left(1-k^2\right)}{d k^2}\right. \\
& =\left(k(k-1)+\frac{k^2-1}{k}\right)^2-\left(\frac{1-k^2}{k}+\frac{1-k}{k^2}\right)^2 \\
& =\left(\frac{k^2(k-1)+\left(k^2-1\right)}{k}\right)^2-\left(\frac{k\left(1-k^2\right)+1-k}{k^2}\right)^2 \\
& =\frac{\left(k^3-1\right)^2}{k^2}-\frac{\left(-k^3+1\right)^2}{k^4}
\end{aligned}
$
$\begin{aligned} & =\frac{\left(k^3-1\right)^2}{k^2}-\frac{\left(1-k^3\right)^2}{k^4} \\ & =\left(\frac{k^3-1}{k^2}\right)^2\left(1-\frac{1}{k^2}\right) \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & \text { R.H.S. }=(a-d)^2\left(\frac{1}{c^2}-\frac{1}{b^2}\right) \\ & =\left(d k^3-d\right)^2\left(\frac{1}{d^2 k^2}-\frac{1}{d^2 k^4}\right) \\ & =d^2\left(k^3-1\right)^2\left(\frac{k^2-1}{d^2 k^4}\right) \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$

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