Question

determinati primul termen si ratia unei progresii aritmetice: a1+a3=10, S6=48

Answer 1

[tex]a_1+a_2+a_3+...+a_6=48 \\ \\ S6=\frac{2a_1+(6-1)d}{2} * 6= (2a_1+5d)*3 \\ (2a_1+5d)*3=48 \\ \\ \boxed{2a_1+5d=16} \\ \\ a_1+a_3=10 ; \ a_3=a_1+2d \\a_1+a_1+2d=10 \\ 2a_1+2d=10 |:2 \\ \\ \boxed{a_1+d=5} \\ \\ a_1= 5-d \\ \\2(5-d)+5d =16 \\ 10-2d +5d=16 \\ 3d=16-10 \\ 3d=6 \\ \\ \boxed{d=2} \\ \\ a_1+d=5 => a_1=5-2 \\ \\ \boxed{a_1=3}[/tex]