Question

If $a_1,a_2,a_3,...,a_n$ are in AP, where $a_i>0$ for all $i$ , then value of the expression $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+.....\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

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

Answer 1

Answer: (C)

Let d be the common difference of the given AP then, $a_2-a_1=a_3-a_2=a_4-a_3$ $=...=a_n-a_{n-1}=d$ ?(i) Now, $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}$ $+....+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ ?(ii) On multiplying by the rationalisation factor of every term in numerator and denominator, we get $\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\frac{\sqrt{a_4}-\sqrt{a_3}}{a_4-a_3}$ $+....+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}}$ [using Eq.(i)] $=\frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\sqrt{a_4}-\sqrt{a_3}$ $+...+\sqrt{a_n}-\sqrt{a_{n-1}}]$ $=\frac{1}{d}(\sqrt{a_n}-\sqrt{a_1})$ $=\frac{1}{d}(\sqrt{a_n}-\sqrt{a_1}).\frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}$ $=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}$ $=\frac{n-1}{d(\sqrt{a_n}+\sqrt{a_1})}$ $[\because a_n=a_1+(n-1)d]$ $=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$ $=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$