Solution: We have given 63, 65, 67, …
a = 63
d = a 2 − a 1 = 65 − 63 = 2
n th term of this A.P. = a n = a + (n − 1) d
an = 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n (1)
3, 10, 17, …
a = 3
d = a 2 − a 1 = 10 − 3 = 7
n th term of this A.P. = 3 + (n − 1) 7
an = 3 + 7n − 7
an = 7n − 4 (2)
It is given that, n th term of these A.P.s 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.P.s are equal to each other.
