Question

Let $a_{1}, a_{2}, a_{3}, \ldots a_{49}, a_{50}$, are in arithmetic progression. If $a _{1}=4$ and $a _{50}=144$, then the value of
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{a_{4}}} \ldots .+\frac{1}{\sqrt{ a _{49}}+\sqrt{ a _{50}}}$ is equal to

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

Answer 1

Answer: (B)

Expression
$= \frac{\left[\left(\sqrt{a_{2}}-\sqrt{a_{1}}\right)+\left(\sqrt{a_{3}}-\sqrt{a_{2}}\right)+\ldots .+\left(\sqrt{a_{50}}-\sqrt{a_{49}}\right)\right]}{d}$
where $d =$ common difference of $A.P.$
$=\frac{\sqrt{ a _{50}}-\sqrt{ a _{1}}}{ d }=\frac{\sqrt{144}-\sqrt{4}}{\left(\frac{140}{49}\right)}=\frac{49}{14}$