Question

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

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of these

Answer 1

The correct option is A $A.P.$

$a,b,c$ are in A.P

$\therefore 2b=a+c$

Now,

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

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

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

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

$b-a=c-b$

$2b=a+c$

thus, given square are in $A.P$