1. The second term of an infinite decreasing geometric progression is 3. Find the smallest possible value of the sum $A$ of this progression, given that $A>0$.
Answer: 12.

Solution. Let the first term of the progression be $a$, and the common ratio be $q$. The sum of the progression $A$ is $\frac{a}{1-q}$. From the condition, we have $3=a q$, from which $a=3 / q$. Therefore, we need to find the minimum value of $A=\frac{3}{q(1-q)}$. Note that from the condition it follows: $0<q<1$. The number $A$ takes its minimum value when the denominator of the fraction $q(1-q)$ is maximized, but $\left(-q^{2}+q\right)$ has a maximum point at $q=1 / 2$, and the maximum value of $q(1-q)$ is $1 / 4$. Then the minimum possible value of $A=3 / 1 / 4=12$.

Comment. The sum $A$ is represented as a function of one variable - 3 points.