Let a_1 , a_2 , a_3 , ... be an arithmethic progression of positive real numbers. Then (1/( (√a_1 )+(√a_2 )))+(1/( (√a_2 )+(√a_3 )))+∙∙∙+(1/( (√a_(n−1) )+(√a_n )))= (A) ((n+1)/( (√a_1 )+(√a_n ))) (B) ((n−1)/( (√a_1 )+(√a_n ))) (C) (n/( (√a_1 )+(√a_n ))) (D) (n/( (√a_n )−(√a


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

$Let a_{1}, a_{2}, a_{3}, . . . be an arithmethic progression of$ $positive real numbers . Then$ $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \cdot \cdot \cdot + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$ $(A) \frac{n + 1}{\sqrt{a_{1}} + \sqrt{a_{n}}} (B) \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$ $(C) \frac{n}{\sqrt{a_{1}} + \sqrt{a_{n}}} (D) \frac{n}{\sqrt{a_{n}} - \sqrt{a_{1}}}$
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

	$a_{n} = a_{1} + (n - 1) d .$ $Thank you, Sir$
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