Question

Intr-o progresie aritmetică avem a2=7 și a10=15 Calculați a2008

Answer 1

$\displaystyle a_2=7,~a_{10}=15,~a_{2008}=? \\ \\ a_2=7 \Rightarrow a_{2-1}+r=7 \Rightarrow a_1+r=7 \Rightarrow a_1=7-r \\ \\ a_{10}=15 \Rightarrow a_{10-1}+r=15 \Rightarrow a_9+r=15 \Rightarrow a_1+9r=15 \Rightarrow \\ \\ \Rightarrow 7-r+9r=15 \Rightarrow -r+9r=15-7 \Rightarrow 8r=8 \Rightarrow r= \frac{8}{8} \Rightarrow r=1\\ \\ a_1=7-r \Rightarrow a_1=7-1 \Rightarrow a_1=6 \\ \\ a_{2008}=a_{2008-1}+r=a_{2007}+r=a_1+2007r=6+2007 \cdot 1= \\ \\ =6+2007=2013\\ \\ \boxed{a_{2008}=2013}$