Câu 9: \(u_8=u_1+7d\) =>\(7d=26-\frac13=\frac{77}{3}\) =>\(d=\frac{11}{3}\) =>Chọn ACâu 10: \(S_{n}=\frac{n\cdot\left\lbrack2u_1+\left(n-1\right)\cdot d\right\rbrack}{2}\) =>\(n\cdot\frac{\left\lbrack2\cdot3+4\left(n-1\right)\right\rbrack}{2}=253\) =>n(3+2n-2)=253=>n(2n+1)=253=>\(2n^2+n-253=0\) =>\(2n^2-22n+23n-253=0\) =>(n-11)(2n+23)=0=>\(\left[\begin{array}{l}n=11\left(nhận\right)\\ n=-\frac{23}{2}\left(loại\right)\end{array}\right.\) =>n=11=>Chọn BCâu 11: ACâu 12: \(u_6=u_1+5d\) =>\(5d=27-\left(-3\right)=27+3=30\) =>d=6=>Chọn DCâu trả lời: Câu 9: \(u_8=u_1+7d\) =>\(7d=26-\frac13=\frac{77}{3}\) =>\(d=\frac{11}{3}\) =>Chọn ACâu 10: \(S_{n}=\frac{n\cdot\left\lbrack2u_1+\left(n-1\right)\cdot d\right\rbrack}{2}\) =>\(n\cdot\frac{\left\lbrack2\cdot3+4\left(n-1\right)\right\rbrack}{2}=253\) =>n(3+2n-2)=253=>n(2n+1)=253=>\(2n^2+n-253=0\) =>\(2n^2-22n+23n-253=0\) =>(n-11)(2n+23)=0=>\(\left[\begin{array}{l}n=11\left(nhận\right)\\ n=-\frac{23}{2}\left(loại\right)\end{array}\right.\) =>n=11=>Chọn BCâu 11: ACâu 12: \(u_6=u_1+5d\) =>\(5d=27-\left(-3\right)=27+3=30\) =>d=6=>Chọn D