A2 Let $x$ be the average of the following six numbers: $\{12,412,812,1212,1612,2012\}$. Determine the value of $x$.
Solution: The answer is $x=1012$.

Solution 1: The sum of the first and sixth terms is 2024. The sum of the second and fifth terms is 2024 and the sum of the third and fourth terms is 2024 . Hence, the sum of the six terms is $2024 \times 3$. Hence, the average of the six terms is
$$
\frac{2024 \times 3}{6}=\frac{2024}{2}=1012 \text {. }
$$

Solution 2: The average of the six numbers is
$$
\begin{array}{l}
\frac{12+412+812+1212+1612+2012}{6} \\
=\frac{0+400+800+1200+1600+2000}{6}+\frac{12+12+12+12+12+12}{6} \\
=\frac{100(4+8+12+16+20)}{6}+12=\frac{100 \times 60}{6}+12=1000+12=1012 \text {. } \\
\end{array}
$$

Solution 3: Note that the sequence is arithmetic. ${ }^{1}$ Therefore, the average of the six numbers is the average of the middle two numbers, which is the halfway point between 812,1212 . Hence, the answer is 1012 .