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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
