Question

Se considera progresia aritmetica (an) n>1 in care a3=5 si a6=11. Sa se calculeze a9.

Answer 1

$\displaystyle a_3=5,~a_6=11,~a_9 =? \\ a_3=5 \Rightarrow a_{3-1}+r=5 \Rightarrow a_2+r=5 \Rightarrow a_1+2r=5 \Rightarrow a_1=5-2r \\ a_6=11 \Rightarrow a_{6-1}+r=11 \Rightarrow a_5+r=11 \Rightarrow a_1+5r=11 \Rightarrow \\ \Rightarrow 5-2r+5r=11 \Rightarrow -2r+5r=11-5 \Rightarrow 3r=6 \Rightarrow r= \frac{6}{3} \Rightarrow r=2 \\ a_1=5-2r \Rightarrow a_1=5-2 \cdot 2 \Rightarrow a_1=5-4 \Rightarrow a_1=1$
$\displaystyle a_9=a_{9-1}+r \Rightarrow a_9=a_8+r \Rightarrow a_9=a_1+8r \Rightarrow a_9= 1+8 \cdot 2 \Rightarrow \\ \Rightarrow a_9=1+16 \Rightarrow \boxed{a_9=17}$