Question

Se considera progresia aritmetica (an)n>/=1.Stiind ca a3+a19=10 , sa se calculeze a6+a16

Answer 1

$\displaystyle a_3+a_{19}=10 ,~a_6+a_{16}=? \\ a_3+a_{19}=10 \Rightarrow a_{3-1}+r+a_{19-1}+r=10 \Rightarrow a_2+r+a_{18}+r=10 \Rightarrow \\ \Rightarrow a_1+2r+a_1+18r=10 \Rightarrow 2a_1+20r=10 \\ a_{6}+a_{16}=a_{6-1}+r+a_{16-1}+r= \\ =a_5+r+a_{15}+r=a_1+5r+a_1+15r=2a_1+20r =10 \\ \boxed{a_6+a_{16}=10}$