數列數學-----奇數項和為44,偶數項為33

設首項 = a
公差 = d
項數 n 為奇數
前n項奇數項和 = a + a+2d + a+4d + a+6d + ... + a+(n-1)d = 44........(1)
前n項偶數項和 = a+d + a+3d + a+5d + a+7d +...+ a+d(n-2)d = 33.......(2)
(1) - (2) :
{ [a - (a+d)] +[ a+2d - (a+3d)] + [a+4d - (a+5d)] + [a+6d - (a+7d)] + ...
+ [a+(n-3)d - ( a+(n-2)d )] } + a+(n-1)d = 44 - 33

==> (-d - d - d - ... - d) + a+(n-1)d = 11
-----------共 (n-1)/2 個 - d-------------
==> (-d)(n-1)/2 + a+(n-1)d = 11
==> a + (n-1)d/2 = 11..........*

又該等差數列前n項和 = [2a + (n-1)d]n/2 = 44 + 33
==> [ a + (n-1)d/2 ] n = 77 , 由 *
==> 11n = 77
n = 7
該等差數列共 7 項
* 成為 a + (7-1)d/2 = 11
==> a + 3d = 11
中間項 = a + 3d = 11
另解 : 奇數項數列中間項 = 奇數項數列和 / 項數 = 77 / 7 = 11


2010-02-05 16:28:50 補充：
兩條題基本一致。

如果各項全是正整數的話,第二題只有一個答案 : 5 7 9 11 13 15 17

中間項是 11 ,共 7 項。

前7項奇數項和 = 5 + 9 + 13 + 17 = 44

前7項偶數項和 = 7 + 11 + 15 = 33

設首項 = a
公差 = d
項數 n 為奇數

前n項奇數項和 = a + a+2d + a+4d + a+6d + ... + a+(n-1)d = 44........(1)

前n項偶數項和 = a+d + a+3d + a+5d + a+7d +...+ a+d(n-2)d = 33.......(2)

(1) - (2) :

{ [a - (a+d)] +[ a+2d - (a+3d)] + [a+4d - (a+5d)] + [a+6d - (a+7d)] + ...

+ [a+(n-3)d - ( a+(n-2)d )] } + a+(n-1)d = 44 - 33


==> (-d - d - d - ... - d) + a+(n-1)d = 11
-----------共 (n-1)/2 個 - d-------------

==> (-d)(n-1)/2 + a+(n-1)d = 11

==> a + (n-1)d/2 = 11..........*


又該等差數列前n項和 = [2a + (n-1)d]n/2 = 44 + 33

==> [ a + (n-1)d/2 ] n = 77 , 由 *

==> 11n = 77

n = 7

該等差數列共 7 項

* 成為 a + (7-1)d/2 = 11

==> a + 3d = 11

中間項 = a + 3d = 11