سوال ۲۸ کنکور ارشد علوم کامپیوتر ۱۴۰۳

This is the arithmetic–geometric mean iteration: $a_{n+1}$ is the geometric mean and $b_{n+1}$ the arithmetic mean of $a_n, b_n$.

$a_n \le b_n$ for all $n$. By AM–GM, $a_{n+1} = \sqrt{a_n b_n} \le \tfrac{a_n + b_n}{2} = b_{n+1}$, and the base case $a_1 < b_1$ is given.

$\{a_n\}$ is increasing. Since $b_n \ge a_n$, $a_{n+1} = \sqrt{a_n b_n} \ge \sqrt{a_n \cdot a_n} = a_n.$

$\{b_n\}$ is decreasing. Since $a_n \le b_n$, $b_{n+1} = \frac{a_n + b_n}{2} \le \frac{b_n + b_n}{2} = b_n.$

So $\{a_n\}$ increases and $\{b_n\}$ decreases (squeezing toward the common AGM limit) — option 2.

Exam trick: the single inequality $\mathrm{GM} \le \mathrm{AM}$ pins down both monotonicities at once.