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Question

If ar>0,rN and a1,a2,a3,...,a2n are in A.P. then a1+a2na1+a2+a2+a2n1a2+a3+a3+a2n2a3+a4+...+an+an+1an+an+1 is equal to

A
n1
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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+1
a1+a2n=a2+a2n1=...=an+an+1=k(say)
Expression =k(a1a2a1a2+...+anan+1anan+1)
=kd(a1an+1), where d= common difference
=kda1an+1a1+an+1=(a1+a2n).ndd(a1+an+1)

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