Solution:

Two APs are given: 63, 65, 67,… and 3, 10, 17,….

Taking first AP,

63, 65, 67, …

The First term, a = 63

The Common difference, d = a₂−a₁ = 65−63 = 2

We all know that the, n^th term of this A.P. = a_n = a+(n−1)d

a_n= 63+(n−1)2 = 63+2n−2

a_n = 61+2n ………………………………………. (i)

Taking second AP,

3, 10, 17, …

The First term, a = 3

The Common difference, d = a₂ − a₁ = 10 − 3 = 7

We all know that,

n^th term of this A.P. = 3+(n−1)7

a_n = 3+7n−7

a_n = 7n−4 ……………………………………………………….. (ii)

Given that the nth term of these A.P.s are equivalent,

When we combine both of these equations, we get:

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

As a result, the 13th terms of both of these A.P.s are equivalent.