Сигурно ти е сбъркано условието.
[tex]\begin{tabular}{|l}a_1+d-3(a_1+2d)=6\\a_1+2d+a_1+3d=-6\\S_n=\frac{2na_1+n(n-1)d}{2}=-20 \end{tabular}[/tex]
[tex]\begin{tabular}{|l}-2a_1-5d=6\\2a_1+5d=-6\\n(2a_1+(n-1)d)=-40 \end{tabular}[/tex]
Сега [tex]n[/tex] е цяло число и проверяваш за всички ..............
[tex]n=2 ; =>d=2;a=11,75[/tex]
[tex]n=3 ; =>d=\frac{22}{9} ; a=-\frac{164}{9}[/tex]
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[tex]n=5 ; =>d=2;a=-8[/tex]
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[tex]n=40 ; =>d=\frac{34}{5} ;a=-20[/tex]
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и така нататък