Question

If a, b, c are in AP , show that

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

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Answer 1

If a, b and c are in AP,

=) b - a = c - b

=) b + b = c + a

=) 2b = a + c - - - - 1)

To show :

$\sf \frac{1}{ \sqrt{b} + \sqrt{c}} , \frac{1}{ \sqrt{c} + \sqrt{a} } , \frac{1}{ \sqrt{a} + \sqrt{b} } \: are \: in \: AP.$ will be in AP.

Then,

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

$\tt if \: \frac{ (\sqrt{b} - \sqrt{a}) }{( \sqrt{c} + \sqrt{a})( \sqrt{b} + \sqrt{c}) } = \frac{ (\sqrt{c} - \sqrt{b} )}{( \sqrt{a} + \sqrt{b})( \sqrt{c} + \sqrt{a}) } . \\ \\ \tt \implies \frac{( \sqrt{b} - \sqrt{a}) }{( \sqrt{b} + \sqrt{c} )} = \frac{( \sqrt{c} - \sqrt{b} )}{( \sqrt{a} + \sqrt{b} )} . \\ \\ \tt \implies \sqrt{b} - \sqrt{a} \times \sqrt{a} + \sqrt{b} = \sqrt{c} - \sqrt{b} \times \sqrt{c} + \sqrt{b} .$

=) b - a = c - b

=) b + b = c + a

=) 2b = a + c,

This is a verification of eq1.

So if a, b and c are in AP, then

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

Answer 2

$\huge \bf{ \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}}$

→ Since a , b , c are in AP , we have

› 2b = (a + c) ..............(i)

Now,

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

will be in AP

$\sf if \: \frac{1}{( \sqrt{c} + \sqrt{a} )} - \frac{1}{ (\sqrt{b} + \sqrt{c} )} \\ = \frac{1}{ (\sqrt{a} + \sqrt{b} )} - \frac{1}{( \sqrt{c} + \sqrt{a} )} \\ \\ \sf \: i.e. \: if \: \frac{ (\sqrt{ b} - \sqrt{a}) }{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c} )} \\ = \frac{( \sqrt{ c} - \sqrt{b} )}{( \sqrt{a} + \sqrt{b} )( \sqrt{c} + \sqrt{a) } } \\ \\ \sf \: i.e. \: if \frac{( \sqrt{b} - \sqrt{a} )}{( \sqrt{b} + \sqrt{c} )} = \frac{( \sqrt{c} - \sqrt{b}) }{( \sqrt{a} + \sqrt{b}) } \\ \\ \sf i.e \: if \: b - a = c - b \\ \\ \sf \: i.e \: if \: 2b \: = a + c \: from \: (i)$

•°• a,b,c are in AP

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

are in AP.