Question

等差数列的前4项和为40.最后4项的和为80.所有各项的和为720.则这个数列共有多少项?

Answer 1

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

Answer 2

等差数列前n项和的定义要理解深刻:
等差数列前n项和=项数*平均数
所以720=n*(40+80)/(4+4)
n=48