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, nth term of this A.P. = an = a+(n−1)d
an= 63+(n−1)2 = 63+2n−2
an = 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,
nth term of this A.P. = 3+(n−1)7
an = 3+7n−7
an = 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.