2. [4 points] Given a finite arithmetic progression $a_{1}, a_{2} \ldots, a_{n}$ with a positive common difference, such that the sum of all its terms is $S$, and $a_{1}>0$. It is known that if the common difference of the progression is tripled, while the first term remains unchanged, then the sum $S$ doubles. By what factor will $S$ increase if the common difference of the original progression is quadrupled (leaving the first term unchanged)?
Answer: It will increase by $\frac{5}{2}$ times.

Solution. The sum of the terms of the original progression is determined by the formula $S_{0}=\frac{2 a_{1}+(n-1) d}{2} \cdot n$, where $d-$ is the common difference of the progression. If $d$ is tripled, the sum of the terms will become $S_{1}=\frac{2 a_{1}+(n-1) 3 d}{2} \cdot n$. According to the condition $S_{1}=2 S_{0}$, we get $2 \cdot \frac{2 a_{1}+(n-1) d}{2} \cdot n=\frac{2 a_{1}+(n-1) 3 d}{2} \cdot n, 4 a_{1}+2(n-1) d=2 a_{1}+3(n-1) d$, $2 a_{1}=(n-1) d$.

Taking this equality into account, $S_{0}=\underline{2 a_{1}+2 a_{1}} \cdot n=2 a_{1} n$. If the common difference $d$ is increased by 4 times, the sum of the terms will become $S_{2}=\frac{2 a_{1}+(n-1) 4 d}{2} \cdot n=\frac{2 a_{1}+8 a_{1}}{2} \cdot n=5 a_{1} n$, which is 2.5 times greater than $S_{0}$.