Question

If $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in A.P., where $a_{i}>0$ for all $i$, show that $1 /\left(\sqrt{a}_{1}+\sqrt{a}_{2}\right)+1 /\left(\sqrt{a}_{2}+\sqrt{a}_{3}\right)+\ldots . .+1 /\left(\sqrt{a}_{n-1}+\sqrt{a}_{n}\right)=(n-1) /\left(\sqrt{a}_{1}+\sqrt{a}_{n}\right)$

Answer 1

Given that $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in A.P. $\forall a_{i}>0$
$$\begin{aligned} &a_{1}-a_{2}=a_{2}-a_{3}=\ldots=a_{n-1}-a_{n}=-d \text { (constant) } \\ &\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \end{aligned}$$
(rationalizing)
$$\begin{aligned} &=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{a_{1}-a_{2}}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{a_{2}-a_{3}}+\cdots+\frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{a_{n-1}-a_{n}} \\ &=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{-d}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{-d}+\cdots+\frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{-d} \\ &=\frac{1}{-d}\left[\sqrt{a_{1}}-\sqrt{a_{n}}\right] \end{aligned}$$
(rationalizing)
$$\begin{aligned} &=\frac{a_{1}-a_{n}}{-d\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)} \\ &\text { (as } a_{n}=a_{1}+(n-1) d \text { ) } \\ &=\frac{-(n-1) d}{-d\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)} \\ &=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \end{aligned}$$

Answer 2

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Answer 3

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