1. **Problem Statement:** We have a sequence $\{a_n\}$ of real numbers such that the subsequences $\{a_{2n}\}$, $\{a_{2n-1}\}$, and $\{a_{3n}\}$ all converge. We want to show that the entire sequence $\{a_n\}$ converges. 2. **Recall the Sandwich Theorem (Squeeze Theorem):** If $x_n \leq y_n \leq z_n$ for all $n$ beyond some index, and if $\lim_{n \to \infty} x_n = \lim_{n \to \infty} z_n = L$, then $\lim_{n \to \infty} y_n = L$. 3. **Key Idea:** To prove $\{a_n\}$ converges, we need to show that all subsequences converge to the same limit. 4. **Given:** - $\lim_{n \to \infty} a_{2n} = L_1$ - $\lim_{n \to \infty} a_{2n-1} = L_2$ - $\lim_{n \to \infty} a_{3n} = L_3$ 5. **Step:** Since $\{a_{2n}\}$ and $\{a_{2n-1}\}$ are subsequences of $\{a_n\}$, if $L_1 \neq L_2$, the sequence $\{a_n\}$ would not converge. So, we must show $L_1 = L_2$. 6. **Use the subsequence $\{a_{6n}\}$:** Note that $a_{6n}$ is a subsequence of $a_{2n}$ (since $6n$ is even) and also a subsequence of $a_{3n}$ (since $6n$ is a multiple of 3). 7. **Limits of $a_{6n}$:** Since $a_{6n}$ is a subsequence of $a_{2n}$, $\lim_{n \to \infty} a_{6n} = L_1$. Also, since $a_{6n}$ is a subsequence of $a_{3n}$, $\lim_{n \to \infty} a_{6n} = L_3$. Therefore, $L_1 = L_3$. 8. **Similarly, consider $a_{6n-3}$:** This is a subsequence of $a_{2n-1}$ (odd indices) and also a subsequence of $a_{3n}$ (since $6n-3 = 3(2n-1)$). 9. **Limits of $a_{6n-3}$:** Since $a_{6n-3}$ is a subsequence of $a_{2n-1}$, $\lim_{n \to \infty} a_{6n-3} = L_2$. Also, since it is a subsequence of $a_{3n}$, $\lim_{n \to \infty} a_{6n-3} = L_3$. Therefore, $L_2 = L_3$. 10. **From steps 7 and 9:** We have $L_1 = L_3$ and $L_2 = L_3$, so $L_1 = L_2 = L_3 = L$. 11. **Conclusion:** Since the subsequences of even and odd indices converge to the same limit $L$, the entire sequence $\{a_n\}$ converges to $L$. **Final answer:** The sequence $\{a_n\}$ converges and $$\lim_{n \to \infty} a_n = L.$$