等差問題 (急)

T(7) = a + (7 - 1)d = a + 6d = 4
T(21) = a + (21 -1)d = a + 20d = -38

a)
解 (1) & (2)
a + 6d = 4 ----------(1)
a + 20d = -38 ----------(2)
(2) - (1)
14d = -38 - 4
d = -3
a = 4 - 6(-3) = 22

所以
首項 : 22
公差 : -3

b)
通項T(n)
T(n) = 22 + (n - 1)(-3) = 25 - 3n //

c)
T(m) = -89
--> 25 - 3m = -89
-3m = -114
m = 38
第38項

(a) 等差數列的公式:
T(n) = a + (n - 1)d
(其中a為首項，d為公差)

T(7) = a + (7 - 1)d = a + 6d = 4......................(1)
T(21) = a + (21 - 1)d = a + 20d = -38..............(2)
解(1)及(2), 得
首項a = 22
公差 d = -3

(b) T(n) = 22 - 3(n - 1)

(c) T(m) = 22 - 3(m - 1) = -89
3(m - 1) = 111
m = 38