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Ответ:

Объяснение:

[tex]\displaystyle\\\left \{ {{S_4-S_6+2a_5+a_3=0} \atop {S_2+S_3=2}} \right.\\\\S_n=\frac{a_1+a_n}{2} *n;~~a_n=a_1+d(n-1)\\\\S_4=\frac{a_1+a_1+3d}{2}*4=(2a_1+3d)*2=4a_1+6d\\\\ S_6=\frac{a_1+a_1+5d}{2}*6=(2a_1+5d)*3=6a_1+15d\\\\S_2=a_1+a_1+d=2a_1+d\\\\S_3=\frac{a_1+a_1+2d}{2} *3=(a_1+d)*3=3a_1+3d\\\\2a_5=2(a_1+4d)=2a_1+8d\\\\a_3=a_1+2d[/tex]

тогда

[tex]\displaystyle\left \{ {{4a_1+6d-6a_1-15d+2a_1+8d+a_1+2d=0} \atop {2a_1+d+3a_1+3d=2}} \right.[/tex]

[tex]\displaystyle\\\left \{ {{a_1+d=0} \atop {5a_1+4d=2}} \right. ;\left \{ {{d=-a_1} \atop {5a_1-4a_1=2}} \right.[/tex]

Ответ:   [tex]\boxed{a_1=2}[/tex]