1. Given two groups of numerical sequences, each consisting of 15 arithmetic progressions containing 10 terms each. The first terms of the progressions in the first group are $1,2,3, \ldots, 15$, and their differences are respectively $2,4,6, \ldots, 30$. The second group of progressions has the same first terms $1,2,3, \ldots, 15$, but the differences are respectively $1,3,5, \ldots, 29$. Find the ratio of the sum of all elements of the first group to the sum of all elements of the second group.

(12 points)
Solution. Let $a_{1}, \ldots, a_{15}$ be the first terms of the progressions, $d_{1}, \ldots, d_{15}$ be the differences of the first group, and $p_{1}, \ldots, p_{15}$ be the differences of the second group. Then the sums of the first group are $S_{1}=\frac{2 a_{1}+9 d_{1}}{2} \cdot 10=10 a_{1}+45 d_{1}, \ldots, S_{15}=\frac{2 a_{15}+9 d_{15}}{2} \cdot 10=10 a_{15}+45 d_{15}$, their total sum is $S_{1}+\ldots+S_{15}=10 a_{1}+45 d_{1}+\ldots+10 a_{15}+45 d_{15}=10 \cdot\left(a_{1}+\ldots+a_{15}\right)+45 \cdot\left(d_{1}+\ldots+d_{15}\right)=$ $=\frac{a_{1}+a_{15}}{2} \cdot 15 \cdot 10+45 \cdot \frac{d_{1}+d_{15}}{2} \cdot 15=\left(\frac{1+15}{2} \cdot 15\right) \cdot 10+45 \cdot\left(\frac{2+30}{2}\right) \cdot 15=8 \cdot 15 \cdot 10+45 \cdot 16 \cdot 15$

Similarly, the sum of the second group is calculated as

$S_{1}=\frac{2 a_{1}+9 d_{1}}{2} \cdot 10=10 a_{1}+45 d_{1}, \ldots, S_{15}=\frac{2 a_{15}+9 d_{15}}{2} \cdot 10=10 a_{15}+45 d_{15}$, their total sum is

$S_{1}+\ldots+S_{15}=10 a_{1}+45 d_{1}+\ldots+10 a_{15}+45 d_{15}=10 \cdot\left(a_{1}+\ldots+a_{15}\right)+45 \cdot\left(d_{1}+\ldots+d_{15}\right)=$ $=\frac{a_{1}+a_{15}}{2} \cdot 15 \cdot 10+45 \cdot \frac{d_{1}+d_{15}}{2} \cdot 15=\left(\frac{1+15}{2} \cdot 15\right) \cdot 10+45 \cdot\left(\frac{1+29}{2}\right) \cdot 15=8 \cdot 15 \cdot 10+45 \cdot 15 \cdot 15$ Then the ratio of the sums $\frac{8 \cdot 15 \cdot 10+45 \cdot 16 \cdot 15}{8 \cdot 15 \cdot 10+45 \cdot 15 \cdot 15}=\frac{160}{151}$

Answer: $\frac{160}{151}$