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Question

If a1,a2,a3,..........,an are consecutive terms of an increasing A.P. and (12−a1)+(22−a2)+(32−a3)+.......+(n2−an)=(n−1)n(n+1)3, then the value of (a5+a3−a26) is equal to

A
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B
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C
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D
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Solution

The correct option is A 1Given, a1,a2...an are in A.PAlso, (12−a1)+(22−a2)+...+(n2−an)=(n−1)n(n+1)3...(1)⇒(12+22+...n2)−(a1+a2+...+an)=(n−1)n(n+1)3∑n2−n2(a1+an)=n(n−1)(n+1)3 [ sum of AP]n(n+1)(2n+1)6−n(n−1)(n+1)3=n2(a1+an)[∑n2 formula]n+13[2n+12−(n−1)]=a1+an2n+1=a1+an...(2)putting n =1 in (1)⇒12−a1=0⇒a1=1from(2), an=n∴(a5+a3−a26)=(a5+d6)=a66=66=1 [∵a3−a2=d] common diff.option A is correct

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