Gujarati
8. Sequences and Series
hard

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

A

$A.P.$

B

$G.P.$

C

$H.P.$

D

None of these

Solution

(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$

==>$2b = a + c$

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

Standard 11
Mathematics

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