If ar>0,r∈N and a1,a2,a3….a2n are A.P. then a1+a2na1+a2+a2+a2n−1a2+a3+a3+a2n−2a3+a4+…..+an+an+1an+an+1=

a1+a2n=a2+a2n−1......=K (say) a1+a2na1+a2+.......+an+an+1an+an+1=Ka1−a2a1−a2+….+an−an+1an−an+1 =−Kda1−an+1a1+an+1 Where d=a1−a2=…..=an−an−1 =na1+a2na1+an+1

If ar>0,r∈N and a1,a2,a3….a2n are A.P. then a1+a2na1+a2+a2+a2n−1a2+a3+a3+a2n−2a3+a4+…..+an+an+1an+an+1=

a1+a2n=a2+a2n−1......=K (say) a1+a2na1+a2+.......+an+an+1an+an+1=Ka1−a2a1−a2+….+an−an+1an−an+1 =−Kda1−an+1a1+an+1 Where d=a1−a2=…..=an−an−1 =na1+a2na1+an+1

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

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$n - 1$

$\frac{n (a_{1} + a_{2 n})}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$

$1$

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

answer is B.

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Detailed Solution

$a_{1} + a_{2 n} = a_{2} + a_{2 n - 1} . . . . . . = K (say)$

$\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} = K \{\frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{a_{1} - a_{2}} + \dots . + \frac{\sqrt{a_{n}} - \sqrt{a_{n + 1}}}{a_{n} - a_{n + 1}}\} = \frac{- K}{d} \frac{a_{1} - a_{n + 1}}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}} (Where d = a_{1} - a_{2} = \dots .. = a_{n} - a_{n - 1}) = \frac{n (a_{1} + a_{2 n})}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$