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Question

If the sum of first n terms of an A.P. is n23, then the sum of squares of first 18 terms is

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Solution

Sn=n23
Sn1=(n1)23
tn=SnSn1 =2n13

Now,
nn=1t2n=(13)2[12+32+...+(2n1)2]
=13[{12+22+...+(2n)2}{22+42+...+(2n)2}]
=13[(2n)(2n+1)(4n+1)622(12+22+...+n2)]
=13[(2n)(2n+1)(4n+1)6(4n)(n+1)(2n+1)6]
=13×2n+13[n(4n+1)2n(n+1)]
=n(4n21)9

For n=18
18n=1t2n=18(4×1821)9=2590

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