等差数列{an}的前n项和为Sn,若a2+a8+a11=30,那么S13=_.

问题描述:

等差数列{an}的前n项和为Sn,若a2+a8+a11=30,那么S13=______.

由等差数列的性质可得:a2+a11=a6+a7
故a2+a8+a11=a6+a7+a8=30,即3a7=30,a7=10,
故S13=

13(a1+a13)
2
=
13×2a7
2
=130
故答案为:130