Question

a, b, c, d are in continued proportion prove that ( b+c) (b+d) =(c+a) (c+d) ​

Answer 1

Answer:

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,

$= (b + c)(b + d)$

$= (d {k}^{2} + dk)(d {k}^{2} + d)$

$= dk(k + 1) \times d( {k}^{2} + 1)$

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

R. H. S,

$= (c + a)(c + d)$

$= (dk + d {k}^{3} )(dk + d)$

$= dk(1 + {k}^{2} ) \times d(k + 1)$

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

therefore,

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