Question

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

Answer 1

Din an => a1+(n-1)*r 
=>a3=a1+2r=5 
si 
a6=a1+5r=11,
 de unde a1=1 
si r=2. 
=> Atunci a9=17

Answer 2

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

$\displaystyle a_9=a_{9-1}+r \Rightarrow a_9=a_8+r \Rightarrow a_9=a_1+8r \Rightarrow a_9=1+8 \cdot 2 \Rightarrow \\ \\ \Rightarrow a_9=1+16 \Rightarrow \boxed{a_9=17}$