If a,b,c are in A.P., then frac{1}{{sqrt a + | vedclass

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Question

(a) $a, b, c$ are in $A.P.$   $i.e.$, $2b = a + c$

Let $\frac{1}{{\sqrt a + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt b }} = \frac{1}{{\sqrt

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$A.P.$

Vedclass · Answer

(a) $a, b, c$ are in $A.P.$   $i.e.$, $2b = a + c$

Let $\frac{1}{{\sqrt a + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt b }} = \frac{1}{{\sqrt b + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt c }}$

==> $\frac{{\sqrt b - \sqrt c }}{{\sqrt a + \sqrt b }} = \frac{{\sqrt a - \sqrt b }}{{\sqrt b + \sqrt c }}$

$ \Rightarrow b - c = a - b$

==>$2b = a + c$

$	herefore \frac{1}{{\sqrt a + \sqrt b }},\frac{1}{{\sqrt a + \sqrt c }},\frac{1}{{\sqrt b + \sqrt c }}$ are in $A.P.$

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

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