问题
选择题
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
答案
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=
=13a7=13×3=39.13(a1+a13) 2
故选D
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=
=13a7=13×3=39.13(a1+a13) 2
故选D