Question

Intr-o progresie aritmetica a4=7 și a9=22.Calculati a14

Answer 1

$\left \{ {{a_4=7} \atop {a_9=22}} \right. \Leftrightarrow \left \{ {{a_1+3r=7} \atop {a_1+8r=22}}. \right. \\ \\ (a_1+8r)-(a_1+3r)=22-7 \Leftrightarrow 5r=15 \Rightarrow r=3. \\ \\ a_1+3r=7 \Rightarrow a_1=7-3r=7-9=-2. \\ \\ a_{14}=a_1+13r=-2+13 \cdot 3=37.$

Answer 2

[tex]r=a9-a4 5r=22-7 5r=15 r=15/5=3[/tex]

[tex]a4=a1+3r 7=a1+9 7-a1-9=0 -a1-2=0 -a1=2 a1=-2[/tex]

[tex]a14=a1+13r a14=-2+13*3=-2+39=37[/tex]