Given that Sn=2n2+n. For Sn−1, we will put (n−1) at the place of n. ⇒Tn=Sn−Sn−1=(2n2+n)−{2(n−1)2+(n−1)} ⇒Tn=2n2+n−{2(n2−2n+1)+n−1} ⇒Tn=2n2+n−{2n2−4n+2+n−1} ⇒Tn=2n2+n−2n2+3n−2+1 ⇒Tn=4n−1 Therefore, T10=4×10−1=39 Therefore, option (3) is the correct answer