等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?

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等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?等差数列{a}的前n项和为Sn,

等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?
等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?

等差数列{a}的前n项和为Sn,S4=20,S(n-4)=60,Sn=120,n=?
由条件有a1+a4=2a1+3d=10;(2a1+(n-5)d)(n-4)=120;(2a1+(n-1)d)n=240;对这个方程组消去a1和d求得n=12;其实也可求出d=5/4,a1=25/8.

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