Question

Let $a_1,a_2,a_3,\dots a_{49},a_{50}$, are in arithmetic progression. If $a_1=4$ and $a_{50}=144$, then the value of
$\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}\dots .+\frac{1}{\sqrt{a_{49}}+\sqrt{a_{50}}}$ is equal to

• A. $\frac{14}{13}$
• B. $\frac{49}{14}$
• C. $\frac{50}{13}$
• D. $\frac{48}{17}$

Answer 1

Answer: (B)

Expression
$=\frac{\left[\left(\sqrt{a_2}-\sqrt{a_1}\right)+\left(\sqrt{a_3}-\sqrt{a_2}\right)+\dots .+\left(\sqrt{a_{50}}-\sqrt{a_{49}}\right)\right]}{d}$
where $d=$ common difference of $A.P.$
$=\frac{\sqrt{a_{50}}-\sqrt{a_1}}{d}=\frac{\sqrt{144}-\sqrt{4}}{\left(\frac{140}{49}\right)}=\frac{49}{14}$