Question

salut...se considera progresia aritmetica in care a3=5 si a6=11.sa se calculeze a14=?va rog ajutati.ma!!!

Answer 1

an=a1+(n-1)r
a3=a1+2r=5
a6=a1+5r=11
-a1-2r=-5
 a1+5r=11
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        3r=6, r=2
a1=5-2r=5-4=1
a14=a1+13r=1+26=27

Answer 2

$\displaystyle a_3=5,~a_6=11,~a_{14}=? \\ a_3=5 \Rightarrow a_{3-1}+r=5 \Rightarrow a_2+r=5 \Rightarrow a_1+2r=5 \Rightarrow a_1=5-2r \\ a_6=11 \Rightarrow a_{6-1}+r=11 \Rightarrow a_5+r=11 \Rightarrow a_1+5r=11 \Rightarrow \\ \Rightarrow 5-2r+5r=11 \Rightarrow -2r+5r=11-5 \Rightarrow 3r=6 \Rightarrow r=2 \\ a_1=5-2r \Rightarrow a_1=5-2 \cdot 2 \Rightarrow a_1=5-5 \Rightarrow a_1=1 \\ a_{14}=a_{14-1}+r \Rightarrow a_{14}=a_{13}+r \Rightarrow a_{14}=a_1+13r \Rightarrow$
$\displaystyle \Rightarrow a_{14}=1+13 \cdot 2 \Rightarrow a_{14}=1+26 \Rightarrow \boxed{a_{14}=27}$