Question 5
Ten numbers $a_1,a_2,\ldots,a_{10}$ satisfy
$$\frac{a_1-1}{10}=\frac{a_2-2}{9}=\frac{a_3-3}{8}=\cdots=\frac{a_{10}-10}{1},\qquad a_1+a_2=41.$$
Find $S=a_1-a_2+a_3-a_4+\cdots+a_9-a_{10}$.
$$\frac{a_1-1}{10}=\frac{a_2-2}{9}=\frac{a_3-3}{8}=\cdots=\frac{a_{10}-10}{1},\qquad a_1+a_2=41.$$
Find $S=a_1-a_2+a_3-a_4+\cdots+a_9-a_{10}$.