Question
If $a, b, c, d$ are in continued proportion, prove that $: (a + d)(b + c) – (a + c)(b + d) = (b – c)^2$
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Answer
$a, b, c, d$ are in continued proportion
$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k\ ($say$)$
$ \therefore c=d k^2 b=c k=d k \cdot k=dk^2$
$ a=b k=d k^2 \cdot k=d k^3$
$ \text { L.H.S. }=(a+d)(b+c)-(a+c)(b+d)$
$ =\left(d k^3+d\right)\left(d k^2+d k\right)-\left(d k^3+d k\right)\left(d k^2+d\right)$
$ =d\left(k^3+1\right) d k(k+1)-d k\left(k^2+1\right) d\left(k^2+1\right)$
$ =d^2 k(k+1)\left(k^3+1\right)-d^2 k\left(k^2+1\right)\left(k^2+1\right)$
$ =d^2 k\left[k^4+k^3+k+1-k^4-2 k^2-1\right]$
$ =d^2 k\left[k^3-2 k^2+k\right]$
$ =d^2 k^2\left[k^2-2 k+1\right]$
$ =d^2 k^2(k-1)^2$
$ \text { R.H.S. }=(b-c)^2$
$ =\left(d k^2-d k\right)^2$
$ =d^2 k^2(k-1)^2$
$ \therefore \text { L.H.S. - R.H.S. }$
$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k\ ($say$)$
$ \therefore c=d k^2 b=c k=d k \cdot k=dk^2$
$ a=b k=d k^2 \cdot k=d k^3$
$ \text { L.H.S. }=(a+d)(b+c)-(a+c)(b+d)$
$ =\left(d k^3+d\right)\left(d k^2+d k\right)-\left(d k^3+d k\right)\left(d k^2+d\right)$
$ =d\left(k^3+1\right) d k(k+1)-d k\left(k^2+1\right) d\left(k^2+1\right)$
$ =d^2 k(k+1)\left(k^3+1\right)-d^2 k\left(k^2+1\right)\left(k^2+1\right)$
$ =d^2 k\left[k^4+k^3+k+1-k^4-2 k^2-1\right]$
$ =d^2 k\left[k^3-2 k^2+k\right]$
$ =d^2 k^2\left[k^2-2 k+1\right]$
$ =d^2 k^2(k-1)^2$
$ \text { R.H.S. }=(b-c)^2$
$ =\left(d k^2-d k\right)^2$
$ =d^2 k^2(k-1)^2$
$ \therefore \text { L.H.S. - R.H.S. }$
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