Question

If a,b,c,d are in continued proportion,prove that:(a²-b²) (c²-d²)=(b²-c²)²

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let,

$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \: \: \: (where \: k \: is \: \: a \: \: constant)$

So,

$= > c = dk$

$= > b = ck \: = dk \times k = d {k}^{2}$

$= > a = bk \: = d {k}^{2} \times k \: = d {k}^{3}$

L. H. S,

$= ( {a}^{2} - {b}^{2} )( {c}^{2} - {d}^{2} )$

$= ( { {dk}^{3} )}^{2} - ( { {dk}^{2}) }^{2} \times ( {dk)}^{2} - {d}^{2}$

$= ( {d}^{2} {k}^{6} - {d}^{2} {k}^{4} )( {d}^{2} {k}^{2} - {d}^{2} )$

$= {d}^{2} {k}^{4} ( {k}^{2} - 1) \times {d}^{2} ( {k}^{2} - 1)$

$= {d}^{4} {k}^{4} ( { {k}^{2} - 1) }^{2}$

R. H. S,

$= ( { {b}^{2} - {c}^{2}) }^{2}$

$= (( { { {dk}^{2}) }^{2} - {(dk)}^{2} )}^{2}$

$= ( { {d}^{2} {k}^{4} - {d}^{2} {k}^{2} )}^{2}$

$= ( { {d}^{2} {k}^{2} ( {k}^{2} - 1)) }^{2}$

$= {d}^{4} {k}^{4} ( { {k}^{2} - 1)}^{2}$

L. H. S = R. H. S (proved)