设等差数列{an}前n项和为Sn,若S9=72,则a2+a4+a9=( ) A.12 B.18 C.24 D.36
问题描述:
设等差数列{an}前n项和为Sn,若S9=72,则a2+a4+a9=( )
A. 12
B. 18
C. 24
D. 36
答
∵等差数列{an}前n项和为Sn,S9=72=
=9a5,∴a5=8.9(a1+a9) 2
故 a2+a4+a9=3a1+12d=3a5=24,
故选C.