Given sum of n terms is Sn=50n+2n(n−7)A…(1)
⇒Sn−1=50(n−1)+2(n−1)(n−8)A…(2)
Subtracting (1) and (2), we get
Sn−Sn−1=50n−50(n−1)+2n(n−7)A−2(n−1)(n−8)A
⇒Sn−Sn−1=50(n−n+1)+2A(n(n−7)−(n−1)(n−8))
⇒Sn−Sn−1=50+2A(n2−7n−n2+9n−8)
⇒Tn=Sn−Sn−1=50+A(n−4)
Hence, T1=50−3A and T2=50−2A
⇒d=T2−T1=A and T50=50+46A.