设等差数列{an}的前n项和为Sn,已知a3=24,S11=0.

问题描述:

设等差数列{an}的前n项和为Sn,已知a3=24,S11=0.
(1)求数列{an}的通项公式;(2)求数列的{an}前项和Sn;

S11=11×(a1+a11)÷2=0
推出a1+a11=0
a1=a3-2d a11=a3+8d
代入:a3-2d +a3+8d=0
解得d=-8 a1=40
an=a3+(n-3)d=48-8n
Sn=n(a1+an)÷2
=n(40+48-n)÷2
=44n-1/2n²