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Question

In the sum of first n terms of an A.P. is cn2. then the sum of squares of these n terms is

A
n(4n21)c26
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B
n(4n2+1)c23
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C
n(4n21)c23
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D
n(4n2+1)c26
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Solution

The correct option is C n(4n21)c23
Given that for an A.P,Sn=cn2
Then Tn=SnSn1=cn2c(n1)2=(2n1)c
Sum of squares of n terms of this A.P.
=T2n==(2n1)2.c2=c2[4n24n+n]=c2[4n(n+1)(2n+1)64n(n+1)2+n]=c2n[2(2n2+3n+1)6(n+1)+33]=c2n[4n213]=n(4n21)c23

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