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Question Number 142865 by EnterUsername last updated on 06/Jun/21

Let a_{1}, a_{2}, a_{3}, \dots be an arithmethic progression of

positive real numbers . Then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \cdot \cdot \cdot + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =

(A) \frac{n + 1}{\sqrt{a_{1}} + \sqrt{a_{n}}} (B) \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

(C) \frac{n}{\sqrt{a_{1}} + \sqrt{a_{n}}} (D) \frac{n}{\sqrt{a_{n}} - \sqrt{a_{1}}}

Commented by som(math1967) last updated on 07/Jun/21

(B) \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

Answered by mr W last updated on 06/Jun/21

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \cdot \cdot \cdot + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}

= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + \cdot \cdot \cdot + \frac{\sqrt{a_{n}} - \sqrt{a_{b - 1}}}{a_{n} - a_{n - 1}}

= \frac{\sqrt{a_{n}} - \sqrt{a_{1}}}{d}

= \frac{(\sqrt{a_{n}} - \sqrt{a_{1}}) (n - 1)}{a_{n} - a_{1}}

= \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}

\Rightarrow a n s w e r (B)

Commented by EnterUsername last updated on 06/Jun/21

a_{n} = a_{1} + (n - 1) d .

Thank you, Sir