Question

if a,b,c are in ap then prove 1/√b+√c,1//√c+√a,1/√a+√b are also in ap​

Answer 1

Given -

a, b and c are in A. P

To prove -

$\frac{1}{ \sqrt{b} + \sqrt{c} } \: , \frac{1}{ \sqrt{c} + \sqrt{a} } , \frac{1}{ \sqrt{a} + \sqrt{b} } \: are \: in \: A. P$

i. e

$\frac{1}{ \sqrt{c} + \sqrt{a} } - \frac{1}{ \sqrt{b} + \sqrt{c} } = \frac{1}{ \sqrt{a} + \sqrt{b} } - \frac{1}{ \sqrt{c} + \sqrt{a} }$

Solution -

a, b and c are in A. P

Then,

b-a=c-b...... 1)

Now,

L.H.S

$= \frac{1}{ \sqrt{c} + \sqrt{a} } - \frac{1}{ \sqrt{b} + \sqrt{c} }$

$= \frac{( \sqrt{b} + \sqrt{c} ) - ( \sqrt{c} + \sqrt{a}) }{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c}) }$

$= \frac{ \sqrt{b} + \sqrt{c} - \sqrt{c} - \sqrt{a} }{( \sqrt{c} + \sqrt{a})( \sqrt{b} + \sqrt{c} ) }$

$= \frac{ \sqrt{b} - \sqrt{a} }{( \sqrt{c} + \sqrt{a})( \sqrt{b} + \sqrt{c}) }$

$= \frac{( \sqrt{b} - \sqrt{a} )( \sqrt{b} + \sqrt{a}) }{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c} )( \sqrt{b} + \sqrt{a}) }$

$= \frac{( { \sqrt{b}) }^{2} - ({ \sqrt{a}) }^{2} }{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c} )( \sqrt{b} - \sqrt{a}) }$

$= \frac{b - a}{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c} )( \sqrt{b} + \sqrt{a} )}$

R. H. S

$= \frac{1}{ \sqrt{a} + \sqrt{b} } - \frac{1}{ \sqrt{c} + \sqrt{a} }$

$= \frac{( \sqrt{c} + \sqrt{a}) - ( \sqrt{a} + \sqrt{b}) }{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a}) }$

$= \frac{ \sqrt{c} + \sqrt{a} - \sqrt{a} - \sqrt{b} }{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a} ) }$

$= \frac{ \sqrt{c} - \sqrt{b} }{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a}) }$

$= \frac{( \sqrt{c} - \sqrt{b} )( \sqrt{c} + \sqrt{b} ) }{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a})( \sqrt{c} + \sqrt{b} ) }$

$= \frac{( { \sqrt{c} )}^{2} - {( \sqrt{b} )}^{2} }{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a})( \sqrt{c} + \sqrt{b} ) }$

$= \frac{c - b}{( \sqrt{a} + \sqrt{b})( \sqrt{c } + \sqrt{a})( \sqrt{c} + \sqrt{b} )}$

$= \frac{b - a}{( \sqrt{a} + \sqrt{b} )( \sqrt{c} + \sqrt{a} )( \sqrt{c} + \sqrt{b}) } (by \: 1)$

$= \frac{b - a}{( \sqrt{c} + \sqrt{a})( \sqrt{b } + \sqrt{c} )( \sqrt{b} + \sqrt{a} ) }$

L. H. S =R. H. S

Then,

$\frac{1}{ \sqrt{c} + \sqrt{a} } - \frac{1}{ \sqrt{b} + \sqrt{c} } = \frac{1}{ \sqrt{a} + \sqrt{b} } - \frac{1}{ \sqrt{c} + \sqrt{a} }$

Therefore,

$\frac{1}{ \sqrt{b} + \sqrt{c} } , \frac{1}{ \sqrt{c} + \sqrt{a} } , \frac{1}{ \sqrt{a} + \sqrt{b} } \: are \: \: in \: \: A. P$