Question

If $a_1, a_2 , a_3 , a_4$ are in A.P., then $\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{a_{4}}}=$

• A. $\frac{\sqrt{a_{4}}-\sqrt{a_{1}}}{a_{3-a_2}}$
• B. $\frac{a_{4}-a_{1}}{a_{3}-a_{2}}$
• C. $\frac{a_{3}-a_{2}}{\sqrt{a_{4}}-\sqrt{a_{1}}}$
• D. $\frac{a_{1}-a_{4}}{a_{3}-a_{2}}$

Answer 1

Answer: (A)

$\frac{\sqrt{a_{4}}-\sqrt{a_{1}}}{a_{3-a_2}}$

Answer 2

$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{a_{4}}}$
$=\frac{\sqrt{a}_{1}-\sqrt{a_{2}}}{\left(\sqrt{a_{1}}+\sqrt{a_{2}}\right)\left(\sqrt{a_{1}}-\sqrt{a_{2}}\right)}$
$+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{\left(\sqrt{a_{2}}+\sqrt{a_{3}}\right)\left(\sqrt{a_{2}}-\sqrt{a_{3}}\right)}+\frac{\sqrt{a_{3}}-\sqrt{a_{4}}}{\left(\sqrt{a}_{3}+\sqrt{a_{4}}\right)\left(\sqrt{a_{3}}-\sqrt{a_{4}}\right)}$
[by rationalisation]
$=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{a_{1}-a_{2}}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{a_{2}-a_{3}}+\frac{\sqrt{a_{3}}-\sqrt{a_{4}}}{a_{3}-a_{4}}$
$\because a_{1}, a_{2}, a_{3}$ and $a_{4}$ are in AP.
$\therefore a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}$
or
$a_{1}-a_{2}=a_{2}-a_{3}=a_{3}-a_{4}$
Thus, $\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{a_{4}}}$
$=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{a_{1}-a_{2}}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{a_{1}-a_{2}}+\frac{\sqrt{a_{3}}-\sqrt{a_{4}}}{a_{1}-a_{2}}$
$=\frac{\left(\sqrt{a_{1}}-\sqrt{a_{2}}\right)+\left(\sqrt{a_{2}}-\sqrt{a_{3}}\right)+\left(\sqrt{a_{3}}-\sqrt{a_{4}}\right)}{\left(a_{1}-a_{2}\right)}=\frac{-\sqrt{a_{4}}+\sqrt{a_{1}}}{a_{1}-a_{2}}$
$=\frac{\sqrt{a_{4}}-\sqrt{a_{1}}}{a_{2}-a_{1}}=\frac{\sqrt{a_{4}}-\sqrt{a_{1}}}{a_{3}-a_{2}}$