👤

Fie progresia aritmetica(an)n>1 in care a3=5 si a6=11.Calculati a14.

Răspuns :

[tex]a_3=a_1+2r \\ a_6=a_1+5r \\ a_1+2r=5 \\ a_1+5r =11\\ Scadem\ din\ a\ doua\ relatie\ pe\ prima\ si\ obtinem: \\ 3r=6 \\ r=2 \\ a_1=5-4=1 \\ a_{14}=a_1+13r=1+26=27[/tex]