Question

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

Answer 1

Let the n^th term of the given progressions be t_n and T_n, 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 n_^th term is given by
t_n = 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 n_^th term is given by
T_n = A + (n - 1)D
⇒ 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Now, t_n = ​T_n
⇒ 61 + 2n​ = 7n - 4​
⇒ 65 = 5n
⇒ n = 13
Hence, the 13^th terms of the AP's are the same.