問中5數學(等差數列)

首10項之和為240
S(10)=240
10[2a+(10-1)d]/2=240
2a+9d=48--(1)

18項為74
a+(18-1)d=74
a+17d=74--(2)

(1)-2(a):
2a+9d-2a-34d=48-148
-25d=100
d=4

a+17d=74
a+17(4)=74
a=6//
首項=6 ,公差=6//

首18項之和
=S(18)
=(18)[2*6+(18-1)6]/2
=1024//




2008-01-23 23:15:54 補充：
首18項之和 =S(18)=(18)[2*6+(18-1)6]/2=1026//

設首項為a
公差為d
T(1) = a
T(2) = a + d
T(3) = a + 2d
T(10) = a + 9d
T(18) = a + 17d = 74 -----(1)
S(n) = n[2a + (n-1)d]/2
S(10) = 10 [ 2a + (10 - 1)d]/2 = 240
2a + 9d = 48 -----(2)
Sub Eq(2) into Eq(1):
2 (74-17d) +9d = 48
25d = 100
d = 4
from Eq(1):
a + 17(4) = 74
a = 8
S(18) = 18 [ 2a + (18-1)d]/2
= 18[ 2(8) + 17(4)] /2
= 756