CAT 2019 Set-1 | Question: 75

Question

If $a_{1},a_{2}\dots$ are in A.P., then, $\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\dots+\frac{1}{\sqrt{a_{n}}+\sqrt{a_{n+1}}}$ is equal to 

• A. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n-1}}}$
• B. $\frac{n}{\sqrt{a_{1}}+\sqrt{a_{n+1}}}$
• C. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
• D. $\frac{n}{\sqrt{a_{1}}-\sqrt{a_{n+1}}}$

Answer 1

Given that $, \frac{1}{ \sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{ \sqrt{a_{2}} + \sqrt{a_{3}}} + \dots +  \frac{1}{ \sqrt{a_{n}} + \sqrt{a_{n + 1}}} $

Here, do the rationalization, we get

$ \frac{1}{ \sqrt{a_{1}} + \sqrt{a_{2}}} \times \left(\frac{ \sqrt{a_{1}} - \sqrt{a_{2}}}{ \sqrt{a_{1}} - \sqrt{a_{2}}}\right) + \frac{1}{ \sqrt{a_{2}} + \sqrt{a_{3}}} \times  \left(\frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{\sqrt{a_{2}} - \sqrt{a_{3}}}\right)+ \dots + \frac{1}{ \sqrt{a_{n}} + \sqrt{a_{n + 1}}} \times \left(\frac{ \sqrt{a_{n}} - \sqrt{a_{n + 1}}}{  \sqrt{a_{n}} - \sqrt{a_{n + 1}}} \right)$

$ \Rightarrow \frac{  \sqrt{a_{1}} - \sqrt{a_{2}} }{a_{1} – a_{2}} + \frac{  \sqrt{a_{2}} - \sqrt{a_{3}} }{a_{2} – a_{3}} + \dots + \frac{  \sqrt{a_{n}} - \sqrt{a_{n}+1} }{a_{n} – a_{n + 1}} \quad [\because (a + b)(a – b) = a^{2} – b^{2}] $

$ \Rightarrow \frac{ \sqrt{a_{1}} - \sqrt{a_{2}}}{-d} + \frac{ \sqrt{a_{2}} - \sqrt{a_{3}}}{-d} + \dots + \frac{ \sqrt{a_{n}} - \sqrt{a_{n+1}}}{-d} \quad [\because \text{In A.P. series common difference}(d)  = \text{second term – first term}] $

$ \Rightarrow \frac{1}{-d} \left(\sqrt{a_{1}} - \sqrt{a_{2}} + \sqrt{a_{2}} - \sqrt{a_{3}} + \dots + \sqrt{a_{n}} - \sqrt{a_{n+1}}\right) $

$ \Rightarrow \frac{1}{-d} \left( \sqrt{a_{1}} - \sqrt{a_{n+1}}\right) $

Again do the rationalization, we get

$ \frac{1}{-d} \left[ \sqrt{a_{1}} - \sqrt{a_{n+1}} \times \left( \frac{\sqrt{a_{1}} + \sqrt{a_{n}+1}}{ \sqrt{a_{1}} + \sqrt{a_{n + 1}}}\right) \right] $

$ \Rightarrow \frac{1}{-d} \left( \frac{a_{1}-a_{n+1}}{ \sqrt{a_{1}} + \sqrt{a_{n+1}}} \right) $

$ \Rightarrow \frac{1}{-d} \left( \frac{ a_{1}-a_{1} – nd}{ \sqrt{a_{1}} + \sqrt{a_{n + 1}}} \right)  \quad [\because a_{n} = a + (n-1)d, \text{here}\; a = \text{first term}, \; d = \text{common difference},\; n = \text{number of terms},\;a_{n} = n^\text{th}\;\text{term of the A.P. series}]$

$ \Rightarrow \boxed{ \frac{n}{ \sqrt{a_{1}} + \sqrt{a_{n + 1}}}}$

Correct Answer $: \text{B}$