设等差数列{an}前n项和为Sn,若S9=72,则a2加a4加a9=?急

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 22:34:02

设等差数列{an}前n项和为Sn,若S9=72,则a2加a4加a9=?急
设等差数列{an}前n项和为Sn,若S9=72,则a2加a4加a9=?急

设等差数列{an}前n项和为Sn,若S9=72,则a2加a4加a9=?急
S9 = (a1 + a9)×9÷2 = 9(a1 + 4d)
所以 9(a1 + 4d) = 72
所以 a1 + 4d = 8
a2 + a4 + a9
= a1 + d + a1 + 3d + a1 + 8d
= 3a1 + 12d
= 3(a1 + 4d)
= 3×8
= 24

因为 an是等差数列
所以 a1+a9 = a2+a8 = a3+a7 = ... = a5+a5
所以 a5 = S9 / 9 = 8
所以 a2+a4+a9 = a5-3d + a5-d + a5+4d = 3*a5 = 24

S9=9a1+36d=72
a2+a4+a9=3a1+12d=24

S(9)=72
a(5)=72/9=8
a(2)=a(5)-3d=8-3d
a(4)=a(5)-d=8-d
a(9)=a(5)+4d=8+4d
所以:a(2)+a(4)+a(9)=24

因为 S9=72 所以9*(a1+a9)/2=72 所以a1+a9=2a5=16 所以a5=8 故a2+a4+a9=3a5=24

24