For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,... and 3, 10, 17,... are equal?

Let the nth term of the given progressions be tn and Tn, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its nth term is given by
tn = a + (n - 1)d 
⇒ 63 + (n - 1) ⨯ 2
⇒ 61 + 2n

The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its nth term is given by
Tn = A + (n - 1)D 
⇒ 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Now, tn = ​Tn
 ⇒ 61 + 2n​ = 7n - 4​
⇒ 65 = 5n
⇒ n = 13
Hence, the 13th terms of the AP's are the same.