63,65,67,……a=63d=a2−a1=65−63=2nth term of this A.P,an=a+(n−1)dan=63+(n−1)2=63+2n−2an=61+2n……(i)3,10,17……a=3d=a2−a1=10−3=7nth term of this A.P=3+(n−1)7an=3+7n−7an=7n−4……(ii)
It is given that, nth term of these A.Ps are equal to each other.
Equating both these equations, we obtain,
61 + 2n = 7n - 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.Ps are equal to each other.