(An),a3=25,a10=-3 найти a,d арифммтическая прогресия (An),a1=7,a3=11 найти a18 s18

Решите задачу:

$a _{3} =25 \\ a _{10} =-3 \\ a _{3} =a _{1} +2d\\a _{10} =a _{1} +9d\\ \\ a _{1} +2d=25 \\ a _{1} +9d=-3/*(-1) \\ \\ a _{1} +2d=25 \\ -a _{1} -9d=3\\ -7d=28\\d=-4 \\ a _{1} -8=25\\ a _{1} =25+8=33$

$a _{1} =7 \\ a _{3} =11 \\ a _{3} =a _{1} +2d \\ 11=7+2d\\2d+7=11\\2d=11-7\\2d=4\\d=2 \\ a _{18} =a _{1} +17d=7+17*2=7+34=41 \\ S _{18} = \frac{2*a _{1} +17d}{2} = \frac{14+34}{2} = \frac{48}{2} =24$