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Question

# If ar>0,rϵN and a1, a2, a3,..., a2n are in AP thena1+a2n√a1+√a2+a2+a2n−1√a2+√a3+a3+a2n−2√a3+√a4+...+an+an+1√an+√an+1is equal to

A
n-1
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B
n(a1+a2n)a1+an+1
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C
n1a1+an+1
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D
none of these
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Solution

## The correct option is B n(a1+a2n)√a1+√an+1Given a1,a2,a3......,a2n are in A.PWe know that Sn=n2(a1+a2n)=n2(a2+a2n−1) and so onwhere Sn is the sum of the given A.PHence each numerator term can be written as 2SnnNow let's multiply the numerator and denominator by the common difference ′d′So the sequence =1d×{(a1+a2n)d√a1+√a2+(a2+a2n−1)d√a2+√a3+.....+(an+an+)d√an+√an+1}=1d×{(a1+a2n)(a2−a1)√a1+√a2+(a2+a2n−1)(a3−a2)√a2+√a3+.....+(an+an+)(an+1−an)√an+√an+1}Now substitute the value of original numerator in the sequence we get:2Snn×1d×{(a2−a1)√a1+√a2+(a3−a2)√a2+√a3+........+(an+1−an)√an+√an+1}=2Snn×1d×{(√a2−√a1)+(√a3−√a2)+.......+(√an+1−√an)}There will be cancellation of terms. Hence the sequence now becomes :2Snn×1d×(√an+1−√a1)=(a1+a2n)×n×(√an+1−√a1)nd=n(a1+a2n)×(√an+1−√a1)(an+1−a1)Here we have written nd=(an+1−a1)Since an+1=a1+nd=n(a1+a2n)×1(√an+1+√a1)=n(a1+a2n)√an+1+√a1

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