在等差数列{an}中,已知a3+a7=20,则a1+2a7=

来源:百度知道 编辑:UC知道 时间:2024/06/06 04:41:00

等差数列:

an = a1 + (n - 1)d
n = 项数
d = 公差

a3 = a1 + 2d
a7 = a1 + 6d

a3 + a7 = 2a1 + 8d = 20

a1 + 2a7 = a1 + 2(a1 + 6d)
= 3a1 + 12d
= 1.5 * (2a1 + 8d)
= 1.5 * 20
= 30

a3+a7=20 a1+2n+a1+6n=20 2a1+8n=20 a1+4n=10 a1+2a1+12n=3a1+12n=30

由等数列性质得:2a5=a3+a7=20,所以a5=10
又a1+2a7=a1+a7+a7=3a5=3*10=30
所以a1+2a7=30

(注:左右下标和相等,则它们的和也相等)

30