已知等差数列an的前n项和为Sn,且a1+2a7+a8+a12=15,则S13=(  ) A.104 B.78 C.52 D.39

问题描述:

已知等差数列an的前n项和为Sn,且a1+2a7+a8+a12=15,则S13=(  )
A. 104
B. 78
C. 52
D. 39

因为a1+2a7+a8+a12=a1+2(a1+6d)+(a1+7d)+(a1+11d)=15,
即5a1+30d=15,即a7=a1+6d=3,
所以S13=

13(a1+a13
2
=13a7=13×3=39.
故选D