If a1,a2,a3,⋯an are in A.P and ai>0 then 1a1+a2+1a2+a3+1a3+a4+⋯1an−1+an=ka1+anthen k=

=a2−a1a2−a1+a3−a2a3−a2+⋯an−an−1an−an−1[∵a2−a1=a3−a2=⋯=d] =1da2−a1+a3−a2+⋯+an−an−1=1d(an−a1)On rationalizingNr=1d(an−a1)(an+a1) =1d[a1+(n−1)d]−a1(an+a1) =1d(n−1)d(an+a1) ⇒(n−1)an+a1=ka1+an ∴k=n−1

If a1,a2,a3,⋯an are in A.P and ai>0 then 1a1+a2+1a2+a3+1a3+a4+⋯1an−1+an=ka1+anthen k=

=a2−a1a2−a1+a3−a2a3−a2+⋯an−an−1an−an−1[∵a2−a1=a3−a2=⋯=d] =1da2−a1+a3−a2+⋯+an−an−1=1d(an−a1)On rationalizingNr=1d(an−a1)(an+a1) =1d[a1+(n−1)d]−a1(an+a1) =1d(n−1)d(an+a1) ⇒(n−1)an+a1=ka1+an ∴k=n−1

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If $a_{1}, a_{2}, a_{3}, \dots a_{n}$ are in A.P and $a_{i} > 0$ then $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}} + \dots \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{k}{\sqrt{a_{1}} + \sqrt{a_{n}}}$ then $k =$

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$n - 1$

$n + 1$

$1 - n$

answer is B.

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Detailed Solution

$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + \dots \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}} [∵ a_{2} - a_{1} = a_{3} - a_{2} = \dots = d] = \frac{1}{d} [\sqrt{a_{2}} - \sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt{a_{n}} - \sqrt{a_{n - 1}}]$

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

On rationalizing $N_{r}$

$= \frac{1}{d} \frac{(a_{n} - a_{1})}{(\sqrt{a_{n}} + \sqrt{a_{1}})} = \frac{1}{d} \frac{[a_{1} + (n - 1) d] - a_{1}}{(\sqrt{a_{n}} + \sqrt{a_{1}})} = \frac{1}{d} \frac{(n - 1) d}{(\sqrt{a_{n}} + \sqrt{a_{1}})} \Rightarrow \frac{(n - 1)}{\sqrt{a_{n}} + \sqrt{a_{1}}} = \frac{k}{\sqrt{a_{1}} + \sqrt{a_{n}}} ∴ k = n - 1$