Question

If $a,b,c$are in AP, then $\frac{1}{(\sqrt{a}+\sqrt{b})},\frac{1}{(\sqrt{a}+\sqrt{c})},\frac{1}{(\sqrt{b}+\sqrt{c})}$ are in

Answer 1

Step 1. Let $\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathbf{-}\mathit{d}\mathbf{,}\mathbf{}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathbf{,}\mathbf{}\mathit{c}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathbf{+}\mathit{d}$ where $d$ is the common difference.

Given $a,b,c$are in AP

Now,

$\begin{array}{rcl}\frac{1}{(\sqrt{a}+\sqrt{b})}& =& \frac{(b-a)}{(\sqrt{a}+\sqrt{b})d}\\ & =& \frac{(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})}{(\sqrt{a}+\sqrt{b})d}\\ & =& \frac{(\sqrt{b}-\sqrt{a})}{d}....\left(i\right)\end{array}$

$\begin{array}{rcl}\frac{1}{(\sqrt{a}+\sqrt{c})}& =& \frac{\left({\displaystyle \frac{1}{2d}}\right)\left(2d\right)}{(\sqrt{a}+\sqrt{c})}\\ & =& \frac{\left({\displaystyle \frac{1}{2d}}\right)\left(\right(A+d)-(A-d\left)\right)}{(\sqrt{a}+\sqrt{c})}\\ & =& \frac{\left({\displaystyle \frac{1}{2d}}\right)(c-a)}{(\sqrt{a}+\sqrt{c})}\\ & =& \frac{\left({\displaystyle \frac{1}{2d}}\right)(\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a})}{(\sqrt{a}+\sqrt{c})}\\ & =& \frac{(\sqrt{c}-\sqrt{a})}{2d}....\left(ii\right)\end{array}$

$\begin{array}{rcl}\frac{1}{(\sqrt{b}+\surd c)}& =& \frac{\left({\displaystyle \frac{1}{d}}\right)\left(\right(A+d)-A)}{(\sqrt{b}+\sqrt{c})}\\ & =& \frac{\left({\displaystyle \frac{1}{d}}\right)(c-b)}{(\sqrt{b}+\sqrt{c})}\\ & =& \frac{\left({\displaystyle \frac{1}{d}}\right)(\sqrt{c}+\sqrt{b})(\sqrt{c}-\sqrt{b})}{(\sqrt{b}+\sqrt{c})}\\ & =& \frac{(\sqrt{c}-\sqrt{b})}{d}....\left(iii\right)\end{array}$

Step 2. Add equation (i) and (iii), we get

$\begin{array}{rcl}\frac{1}{(\sqrt{a}+\sqrt{b})}+\frac{1}{(\sqrt{b}+\sqrt{c})}& =& \frac{(\sqrt{b}-\sqrt{a})}{d}+\frac{(\sqrt{c}-\sqrt{b})}{d}\\ & =& \frac{(\sqrt{c}-\sqrt{a})}{d}\\ & =& \frac{2}{(\sqrt{a}+\sqrt{c})}(from(ii\left)\right)\end{array}$

Hence, $\frac{\mathbf{}\mathbf{1}}{\mathbf{(}\sqrt{\mathbf{a}}\mathbf{+}\sqrt{\mathbf{b}}\mathbf{)}}\mathbf{,}\mathbf{}\frac{\mathbf{1}}{\mathbf{(}\sqrt{\mathbf{a}}\mathbf{+}\sqrt{\mathbf{c}}\mathbf{)}}\mathbf{,}\mathbf{}\frac{\mathbf{1}}{\mathbf{(}\sqrt{\mathbf{b}}\mathbf{+}\sqrt{\mathbf{c}}\mathbf{)}}$ are in AP.