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 increased by 4 times, while keeping the first term unchanged, then the sum $S$ increases by 3 times. By how many times will $S$ increase if the common difference of the original progression is doubled (leaving the first term unchanged)?
Answer: It will increase by $\frac{5}{3}$ 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 increased by four times, the sum of the terms will become $S_{1}=$ $\frac{2 a_{1}+(n-1) 4 d}{2} \cdot n$. According to the condition $S_{1}=3 S_{0}$, we get $3 \cdot \frac{2 a_{1}+(n-1) d}{2} \cdot n=\frac{2 a_{1}+(n-1) 4 d}{2} \cdot n, 6 a_{1}+3(n-$ 1) $d=2 a_{1}+4(n-1) d, 4 a_{1}=(n-1) d$.

Considering this equality, $S_{0}=\frac{2 a_{1}+4 a_{1}}{2} \cdot n=3 a_{1} n$. If the common difference $d$ is increased by 2 times, the sum of the terms will become $S_{2}=\frac{2 a_{1}+(n-1) 2 d}{2} \cdot n=\frac{2 a_{1}+8 a_{1}}{2} \cdot n=5 a_{1} n$, which is $\frac{5}{3}$ times greater than $S_{0}$.