已知数列an的前n项和为Sn,且满足a2+a4=-7,S6=-24,求数列an的通项公式

2025-06-29 01:10:42
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回答1:

题目应该是等差数列
(a1+d)+(a1+3d)=-7
2(a1+2d)=-7
2a1+4d=-7
6a1+[6(6-1)/2d=-24
2a1+5d=-12
d=-5
a1=13/2
an=a1+(n-1)d
an=-5n+(23/2)