MATHS(等差及等比數列)

已知 a + m , a + 3m和 a + 11m成一個等比數列 , 其中 a 和 m 是不等於 0 的數
(a) 求 m , 答案以 a 表示
Tn = brn-1
T1 = b = a+m ... (1)
T2 = br = a+3m ... (2)
T3 = br2 = a+11m ... (3)
(2)/(1) => r = (a+3m)/(a+m) ... (4)
(3)/(2) => r = (a+11m)/(a+3m) ...(5)
(4) = (5) =>
(a+3m)/(a+m) = (a+11m)/(a+3m)
(a+3m)(a+3m) = (a+11m)(a+m)
a2 + 6am + 9m2 = a2 + 12am + 11m2
2m2 + 6am = 0
2m(m + 3a) = 0
m = 0 (rej) or m = -3a
(b) 由此 , 求該等比數列的公比
代入(4),
公比 r = (a - 9a)/(a - 3a) = (-8a)/(-2a) = 4

13. 已知 a + m , a + 3m和 a + 11m成一個等比數列 , 其中 a 和 m 是不等於 0 的數

(a) 求 m , 答案以 a 表示
T(n)= arn-1
T(1)= a= a+m
T(2)= ar= a+3m
T(3)= ar2= a+11m

T(2) - T(1),
ar - a = (a+3m) - (a+m)
a(r-1) = 2m
m=a(r-1)/2 ________ (4)

Put (4) into T(3),
ar2= a+11m
ar2= a+11a(r-1)/2
r2= 1+11(r-1)/2
2r2= 2+11(r-1)
2r2-11r + 9 =0
(2r-9)(r-1)=0
r=1 or 9/2

Put r=1 into T(2),
ar= a+3m
a = a + 3m
m = 0 (rej)

Put r=9/2 into T(2),
ar= a+3m
9/2 a = a + 3m
7/2 a = 3m
therefore, m = 7/6 a
(b) 由此 , 求該等比數列的公比
T(n)= arn-1
r = 9/2
a = T(1) = a+7/6 a= 13/6 a
therefore,

T(n)= (13/6 a)(9/2)n-1