Maths F5

2008-10-15 2:43 am
求以下等差數列的項數
(13) 7, 11, 15, … , 83

(14) -93, -86, -79 …, 5

回答 (4)

2008-10-15 3:08 am
✔ 最佳答案
(13)
7, 11, 15, … , 83

T(1) = a = 7
d = 11 - 7 = 15 - 11 = 4
T(n) = 83

T(n) = a + (n - 1)d
83 = 7 + (n - 1)4
4(n - 1) = 76
n - 1 = 19
n = 20

項數 = 20

=====
(14)
-93, -86, -79 …, 5

T(1) = a = -93
d = -86 - (-93) = -79 - (-86) = 7
T(n) = 5

T(n) = a + (n - 1)d
5 = -93 + (n - 1)7
7(n - 1) = 98
n - 1 = 14
n = 15

項數 = 15
=
2008-10-15 3:09 am
利用 T(n) = a + (n - 1)d
a)T(n) = 83、a = 7 以及 d = 4
7 + (n - 1)(4) = 83
n = 20
即是第20項

b)T(n) = 5、a = -93 以及 d = 7
-93 + (n - 1)(7) = 5
n = 15
即是第15項
2008-10-15 3:04 am
1.7,11,15,19,23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83

2. -93, -86, -79, -62 , -41, -16 , 5
參考: 自己
2008-10-15 3:04 am
13) 1+(n-1)x4
=1+4n-4
=4n-3
14) -93+(n-1)x7
=-93+7n-7
=7n-100


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