If the  sum of  first  n terms  of  an  AP  is cn2,  then the sum of squares of these n terms  is n4n2−1c2k then k is

Sn=cn2Sn−1=c(n−1)2=cn2+c−2cnTn=2cn−cTn2=(2cn−c)2=4c2n2+c2−4c2n Sum =ΣTn2=4c2+n(n+1)(2n+1)6+nc2−2c2n(n+1)=2c2n(n+1)(2n+1)+3nc2−6c2n(n+1)3=nc24n2+6n+2+3−6n−63=nc24n2−13