Given that,S7=49S17=289S7=72[2a+(n−1)d]49=72[2a+(7−1)d]7=(a+3d)a+3d=7……(i)Similarly,S17=172[2a+(17−1)d]289=172(2a+16d)17=(a+8d)a+8d=17……(ii)Subtracting equation (i) from equation (ii),5d=10d=2From equation (i),a+3(2)=7a+6=7a=1Sn=n2[2a+(n−1)d]=n2[2(1)+(n−1)×2]=n2(2+2n−2)=n2(2n)=n2