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Question

If a1,a2,a3,..........,an are consecutive terms of an increasing A.P. and (12a1)+(22a2)+(32a3)+.......+(n2an)=(n1)n(n+1)3, then the value of (a5+a3a26) is equal to

A
1
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B
1
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C
2
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D
2
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Solution

The correct option is A 1
Given, a1,a2...an are in A.P
Also, (12a1)+(22a2)+...+(n2an)=(n1)n(n+1)3...(1)
(12+22+...n2)(a1+a2+...+an)=(n1)n(n+1)3
n2n2(a1+an)=n(n1)(n+1)3 [ sum of AP]
n(n+1)(2n+1)6n(n1)(n+1)3=n2(a1+an)[n2 formula]
n+13[2n+12(n1)]=a1+an2
n+1=a1+an...(2)
putting n =1 in (1)12a1=0a1=1
from(2), an=n
(a5+a3a26)=(a5+d6)=a66=66=1 [a3a2=d] common diff.
option A is correct

1136175_1136544_ans_b415209aca0d4a33980d5492c31cc0ff.jpg

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