等差数列的前四项和为40,最后四项和为80,所有各项的和为720,则这一数列共有几项?

题目

等差数列的前四项和为40,最后四项和为80,所有各项的和为720,则这一数列共有几项?

答案

a = a1 + (n-4)d
a = a2 + (n-4)d
a = a3 + (n-4)d
an = a4 + (n-4)d
所以 后四项之和
a + a + a + an
= a1 + a2 + a3 + a4 + 4(n-4)d
= 40 + 4(n-4)d
= 80
得到
(n-4)d = 10
a1 + a2 + a3 + a4 = 40
a1 + (a1 + d) + (a1+2d) + (a1 + 3d) = 40
4a1 + 6d = 40
得到
2a1 = 20 - 3d
所有各项的和为720
a1 + a2 + …… + an = 720
(a1 + an)*n/2 = 720
[a1 + a1 + (n-1)d]*n/2 = 720
[2a1 + (n-1)d]*n/2 = 720
以 2a1 = 20 - 3d 代入上式
[20 - 3d + (n-1)d]*n/2 = 720
[20 + (n-4)d]*n/2 = 720
以 (n-4)d = 10 代入上式
[20 + 10]*n/2 = 720
n = 48