[tex]\displaystyle\bf\\2(3)\\\\c_{1} =33\\\\d=7\\\\\\S_{20} =\frac{2a_{1}+19d }{2} \cdot 20=\Big(2a_{1} +19d\Big)\cdot 10=\Big(2\cdot33+19\cdot7\Big)\cdot 10=\\\\\\=\Big(66+133\Big)\cdot10=199\cdot 10=1990\\\\\\2(4)\\\\c_{1} =-7\\\\d=-6\\\\\\S_{20} =\frac{2a_{1}+19d }{2} \cdot 20=\Big(2a_{1} +19d\Big)\cdot 10=\Big(2\cdot(-7)+19\cdot(-6)\Big)\cdot 10=\\\\\\=\Big(-14-114\Big)\cdot10=-128\cdot 10=-1280[/tex]

[tex]\displaystyle\bf\\4(1)\\\\x_{n} =4n-3\\\\x_{1} =4\cdot 1-3=4-3=1\\\\x_{30} =4\cdot 30-3=120-3=117\\\\\\S_{30} =\frac{x_{1} +x_{30} }{2} \cdot 30=\Big(x_{1} + x_{30} \Big)\cdot 15=\Big(1+117\Big)\cdot 15=\\\\\\=118\cdot 15=1770[/tex]