[tex]\bf 1.\ \ \it 2a+1=\dfrac{a-3+5a+7}{2}\Rightarrow 2a+1=\dfrac{6a+4}{2} \Rightarrow 2a+1=3a+2\Rightarrow a=-1\\ \\ \\ \bf 3.\ \it S_{10}=\dfrac{a_1+a_{10}\cdot10}{2}=\dfrac{(-1+15)\cdot10}{2}=14\cdot5\\ \\ \\ \bf 5.\ \ \it r=\dfrac{a_{10}-a_1}{9}=\dfrac{-2-10}{9}=\dfrac{-12^{(3}}{9}=-\dfrac{4}{3}[/tex]
