Question:

1. Identify the first term and common difference of each AP.
- First AP: $a_1 = 63$, $d_1 = 65-63 = 2$.
- Second AP: $a_2 = 3$, $d_2 = 10-3 = 7$.

2. Write the expression for the nth term of each AP using $a_n = a + (n-1)d$.
- For the first AP: $$a_n = 63 + (n-1) \times 2 = 2n + 61.$$
- For the second AP: $$b_n = 3 + (n-1) \times 7 = 7n - 4.$$

3. Set the two nth terms equal because we are asked when they are the same.
$$2n + 61 = 7n - 4.$$

4. Solve the linear equation for $n$.
\[\begin{aligned}
2n + 61 &= 7n - 4 \\
61 + 4 &= 7n - 2n \\
65 &= 5n \\
n &= \frac{65}{5} = 13.
\end{aligned}\]

5. Verify by substituting $n = 13$ back into both nth‑term formulas.
- First AP: $a_{13} = 2(13) + 61 = 87$.
- Second AP: $b_{13} = 7(13) - 4 = 87$.
Both give the same value, confirming the solution.