Question
If $a, b, c, d$ are in continued proportion$,$ prove that$: (a^2 – b^2) (c^2 – d^2) = (b^2 – c^2)^2$
✓
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^{\prime} b = ck = dk \cdot k = dk ^2,$
$ a = bk = dk ^2 \cdot k = dk ^3$
$ \text { L.H.S. }=\left( a ^2- b ^2\right)\left( c ^2- d ^2\right)$
$ =\left[\left( dk ^3\right)^2-\left( dk ^2\right)^2\right]\left[( dk )^2- d ^2\right]$
$ =\left( d ^2 k ^6- d ^2 k ^4\right)\left( d ^2 k ^2- d ^2\right)$
$ = d ^2 k ^4\left( k ^2-1\right) d ^2\left( k ^2-1\right)$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \text { R.H.S. }=\left(b^2- c ^2\right)^2$
$ =\left[\left( dk ^2\right)^2-( dk )^2\right]^2$
$ =\left[d^2 k ^2- d ^2 k ^2\right]^2$
$ =\left[ d ^2 k ^2\left( k ^2-1\right)\right]^2$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k\ ($say$)$
$ \therefore c=d k^{\prime} b = ck = dk \cdot k = dk ^2,$
$ a = bk = dk ^2 \cdot k = dk ^3$
$ \text { L.H.S. }=\left( a ^2- b ^2\right)\left( c ^2- d ^2\right)$
$ =\left[\left( dk ^3\right)^2-\left( dk ^2\right)^2\right]\left[( dk )^2- d ^2\right]$
$ =\left( d ^2 k ^6- d ^2 k ^4\right)\left( d ^2 k ^2- d ^2\right)$
$ = d ^2 k ^4\left( k ^2-1\right) d ^2\left( k ^2-1\right)$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \text { R.H.S. }=\left(b^2- c ^2\right)^2$
$ =\left[\left( dk ^2\right)^2-( dk )^2\right]^2$
$ =\left[d^2 k ^2- d ^2 k ^2\right]^2$
$ =\left[ d ^2 k ^2\left( k ^2-1\right)\right]^2$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
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