Question

a4+a8=30...10a1-4a7=-45...a1;r=?

Answer 1

$\displaystyle \left \{ {{a_4+a_8=30} \atop {10a_1-4a_7=-45}} \right. \Rightarrow \left \{ {{a_{4-1}+r+a_{8-1}+r=30} \atop {10a_1-4(a_{7-1}+r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_3+r+a_7+r=30} \atop {10a_1-4(a_6+r)=-45 }} \right. \Rightarrow \left \{ {{a_1+3r+a_1+7r=30} \atop {10a_1-4(a_1+6r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{2a_1+10r=30} \atop {10a_1-4a_1-24r=-45}} \right. \Rightarrow \left \{ {{2a_1+10r=30/\cdot(-3)} \atop {6a_1-24r=-45}} \right. \Rightarrow$
$\displaystyle \Rightarrow \left \{ {{-6a_1-30r=-90} \atop {6a_1-24r=-45}} \right. \\ ~~~~----------- \\ ~~~~~~~~~/ ~~~-54r=-135 \Rightarrow r= \frac{135}{54} \Rightarrow \boxed{r= \frac{5}{2}} \\ 2a_1+10r=30 \Rightarrow 2a_1+10 \cdot \frac{5}{2} =30 \Rightarrow 2a_1+ \frac{50}{2} =30 \Rightarrow \\ \Rightarrow 2a_1+25=30 \Rightarrow 2a_1=30-25 \Rightarrow 2a_1=5 \Rightarrow \boxed{a_1= \frac{5}{2}}$