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=______.
答
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
=13013×2a7
2
故答案为:130