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

Answer 1

The correct option is C A.P.

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

Let the common difference be $k$.

$\Rightarrow b=a+k,c=b+k$...(1)

Now the terms $\frac{1}{\sqrt{b}+\sqrt{c}}$, $\frac{1}{\sqrt{c}+\sqrt{a}}$, $\frac{1}{\sqrt{a}+\sqrt{b}}$ can also be written as

$\frac{\sqrt{b}-\sqrt{c}}{b-c},\frac{\sqrt{c}-\sqrt{a}}{c-a},\frac{\sqrt{a}-\sqrt{b}}{a-b}$ (rationalizing the denominators)

Using (1), these terms can be written as

$\frac{\sqrt{b}-\sqrt{c}}{-k},\frac{\sqrt{c}-\sqrt{a}}{2k},\frac{\sqrt{a}-\sqrt{b}}{-k}$

Let these terms be $t_1,t_2,t_3$ respectively.

$t_1+t_3=\frac{\sqrt{b}-\sqrt{c}}{-k}+\frac{\sqrt{a}-\sqrt{b}}{-k}$

$=\frac{\sqrt{c}-\sqrt{a}}{k}=2t_2$

Hence, $t_1,t_2,t_3$ are in A.P.