Question

If $a_1,a_2,......a_n$are in arithmetic progression, where $a_i>0$ for all $i$. Then, $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+.............+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

• A. $n^2(n+1)/2$
• B. $\frac{(n-1)}{\sqrt{a_1}+\sqrt{a_n}}$
• C. $n(n-1)/2$
• D. None of these

Answer 1

The correct option is B

$\frac{(n-1)}{\sqrt{a_1}+\sqrt{a_n}}$

Explanation for the correct option:

Calculate the value of $\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}_{\mathbf{1}}}\mathbf{}\mathbf{+}\sqrt{{\mathbf{a}}_{\mathbf{2}}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\sqrt{{\mathbf{a}}_{\mathbf{3}}}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}}\mathbf{}\mathbf{+}\sqrt{{\mathbf{a}}_{\mathbf{n}}}}$:

Here $a_1,a_2,\dots ..,a_n$ are in AP

$\begin{array}{rcl}{a}_1-a_2& =& a_2-a_3=\dots .=a_{(n-1)}-a_n=-d\\ & & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+.........+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\\ & & \\ & & \mathrm{By}\mathrm{rationalization}\\ & & \\ & =& \frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}+\frac{\sqrt{a_2}-\sqrt{a_3}}{a_2-a_3}+.........+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-a_n}\\ & =& \frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+.........+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}\\ & =& \left(\frac{1}{-d}\right)\left(\sqrt{a_1}-\sqrt{a_n}\right)\end{array}$

Multiply numerator and denominator with $(\sqrt{a_1}+\sqrt{a_n})$

$\begin{array}{rcl}& =& \frac{(a_1-a_n)}{(-d)\left(\sqrt{a_1}+\sqrt{a_n}\right)}\\ & =& \frac{\left[a_1-a_1-(n-1)d\right]}{(-d)\left(\sqrt{a_1}+\sqrt{a_n}\right)}\\ & =& \frac{-(n-1)d}{(-d)\left(\sqrt{a_1}+\sqrt{a_n}\right)}\\ & =& \frac{(n-1)}{\left(\sqrt{a_1}+\sqrt{a_n}\right)}\end{array}$

Hence, the correct option is (B).