Geometric Sequence part 2

2012-08-31 7:24 pm

1. the sum to infinity of a geometric sequence T1,T2,T3,T4,... is 6 and T1+T3+T5+T7+...=9.

a) find the common ratio of the sequence

b) find T1.

2. The sum of the first 3 terms of a geometric sequence is 13/2, and the sum of the 4th,5th and 6th terms is -13/128.

a) find the common ratio of the sequence
b)find the first term of the sequence
c) find the sum to infinity of the sequence
d) at least how many successive negative terms should be added together to make their sum smaller than -2.133?

3a) find the general term of the geometric sequence 16,48,144,...
b) determine whether log16,log48,log144,... is a gemoetric sequence

更新1:

大師兄,第3題個ANSWER係YES,而唔係NO,唔知係答案錯定係點呢?? 我都計到NO

回答 (1)

myisland8132

2012-09-01 6:28 pm

✔ 最佳答案

1(a) a/(1 - r) = 6a/(1 - r^2) = 96/(1 + r) = 9 => r = -1/3(b) T1 = 82(a) a + ar + ar^2 = 13/2ar^3 + ar^4 + ar^5 = -13/128ar^3(1 + r + r^2) = -13/128(13/2)r^3 = -13/128r = -1/4(b) a(1 + r + r^2) = 13/2a(13/16) = 13/2a = 8(c) S = a/(1 - r) = 8/(1 + 1/4) = 32/5(d) Consider -2, -1/8, ...-(2 + 1/8 + 1/128 + 1/2048) = 2.13330078 So, 4 terms is need3(a) 16 * 3^(n - 1)(b) log48/log16 1.39624063log144/log48 = 1.28379107So, log16,log48,log144 does not form a geometric sequence

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