f.5 maths 等比數列 ! 急

Q1. Formula of general term is T(n) = ar^(n -1).
A) Comparing with the formula, a = 4 and r = 5.
B) 3(2)^n = 3(2)^(n -1)(2) = 6(2)^(n -1). So a = 6 and r = 2.
C) T(1) = 2^(-1) = 1/2. T(2) = 2^(-2) = 1/4. So a = 1/2 and r = (1/4)/(1/2)
= 1/2.
D) T(1) = -1/7. T(2) = -1/7^2 = -1/49. So a = -1/7, r = (-1/49)/(-1/7) = 1/7.
Q2.
A) T(n) = (1)(4)^(n - 1) = 4^(n - 1).
B) T(n) = (2)(-2)^(n - 1) = (-1)^(n -1)(2)^(n - 1 + 1) = (-1)^(n -1)2^n
C) T(n) = (-4)(1/4)^(n - 1) = (-1)(4)(4)^(- n + 1) = -1(4)^(2 - n)
Q3.
T(5) = ar^(5 - 1) = ar^4 = 32/3, since r = 2, that is a(2)^4 = 32/3,
16a = 32/3, a = 2/3.
Q4.
a = -3, r = 6/(-3) = -2.
T(k) = (-3)(-2)^(k - 1) = 384
(-2)^(k - 1) = - 128
(-2)^k/(-2) = - 128
(-2)^k = 256 = 2^8, so k = 8.
Q5.
A) a = 1, r = 3, T(n) = (1)(3)^(n - 1) = 3^11, so n - 1 = 11, n = 12.
B) a = -2, r = (-4)/(-2) = 2. T(n) = (-2)(2)^(n - 1) = - 64
2^(n - 1) = 32 = 2^5, n - 1 = 5, so n = 6.
C)a = 7, r = 21/7 = 3. T(n) = 7(3)^(n -1) = 15309
3^(n - 1) = 2187
n - 1 = log2187/log3 = 7, so n = 8.
D) a = 4, r = 8/4 = 2, T(n) = 4(2)^(n -1) = 32768
2^(n - 1) = 8192
n - 1 = log8192/log2 = 13, so n = 14.
Q6.
A) a = 1/8, r = (1/4)/(1/8) = 2, T(k) = (1/8)(2)^(k - 1) = 64
2^(k - 1) = 512 = 2^9, so k - 1 = 9, k = 10.
B) (1/8)(2)^(k - 1) = 4000
2^(k - 1) = 32000
k - 1= log32000/log2 = 14.96, k = 15.96, so k = 16 will be just over 4000.
Q7.
If a, b and c are in G.S., then b^2 = ac.
in this question, that is (a + 3m)^2 = (a+m)(a + 11m)
a^2 + 9m^2 + 6am = a^2 + 11m^2 + 12am
2m^2 + 6am = 0
m^2 + 3am = 0
m(m + 3a) = 0
m + 3a = 0, since m is not equal to zero
so m = -3a.
so common ratio = (a + 3m)/(a + m) = (a - 9a)/(a- 3a) = -8a/-2a = 4.