I sposób
\left \{ {{a_4=2} \atop {a_6=14}} \right.
\left \{ {{a_1+3r=2} \atop {a_1+5r=14 \ \*(-1)}} \right.
\left \{ {{a_1+3r=2} \atop {-a_1-5r=-14}} \right.
-2r=-12 \ |:(-2)
r=6
a_1+3r=2
a_1+3\cdot 6=2
a_1+18=2 \ |-18
a_1=-16
a_3=a_1+2r=-16+2\cdot 6=-16+12=-4
II sposób
\left \{ {{a_4=2} \atop {a_6=14}} \right.
a_4+2r=a_6
2+2r=14
2r=12
r=6
a_3+r=a_4
a_3=a_4-r=2-6=-4
Odpowiedź:
a_3=-4