Question

Se considera progresia aritmetica ( an ) n ≥ 1 in care a2 = -3 si a9 = -24. Sa se calculeze a2011.

Answer 1

$\displaystyle a_2=-3,~a_9=-24,~a_{2011}=? \\ \\ a_2=-3 \Rightarrow a_{2-1}+r=-3 \Rightarrow a_1+r=-3 \Rightarrow a_1=-3-r \\ \\ a_9=-24 \Rightarrow a_{9-1}+r=-24 \Rightarrow a_8+r=-24 \Rightarrow a_1+8r=-24 \Rightarrow \\ \\ \Rightarrow -3-r+8r=-24 \Rightarrow -r+8r=-24+3 \Rightarrow 7r=-21 \Rightarrow \\ \\ \Rightarrow r=- \frac{21}{7} \Rightarrow r=-3 \\ \\ a_1=-3-r \Rightarrow a_1=-3-(-3) \Rightarrow a_1=-3+3 \Rightarrow a_1=0$

$\displaystyle a_{2011}=a_{2011-1}+r \Rightarrow a_{2011}=a_{2010}+r \Rightarrow a_{2011}=a_1+2010r \Rightarrow \\ \\ \Rightarrow a_{2011}=0+2010 \cdot (-3) \Rightarrow a_{2011}=0-6030 \Rightarrow a_{2011}=-6030$