geometric progression anybody?
2018-01-03 5:43 pm
For geometric progression 3,6,12,... find the smallest value of n such that the nth term is greater than 5000.
回答 (5)
2018-01-03 7:47 pm
✔ 最佳答案
First term, a = 3Common ratio, r = 6/3 = 12/6 = 2
T(n) > 5000
a rⁿ⁻¹ > 5000
3 × 2ⁿ⁻¹ > 5000
2ⁿ⁻¹ > 5000/3
log(2ⁿ⁻¹) > log(5000/3)
(n - 1) log(2) > log(5000/3)
n - 1 > log(5000/3)/log(2)
n > 1 + log(5000/3)/log(2)
n > 11.7
The smallest value of n = 12
2018-01-03 7:20 pm
log₂(5000/3) ≈ 10.7
So the first value occurs with n = 12 yielding 3*2¹¹, which is 6144.
So the first value occurs with n = 12 yielding 3*2¹¹, which is 6144.
2018-01-04 12:58 am
#n the term = ar^(n-1)
Where
a = 3
r = 2
Hence
5000 < 3(2^(n- 1))
5000/3 < 2^(n-1)
1666.666... < 2^(n-1)
Take logs
log1666.6666 ... < n - 1log(2)
3.22184... < n(0.30103) - 0.30103
n > = (3.22184 + 0.30103) / 0.30103
n > 10.7027///
nth term is 11.
Where
a = 3
r = 2
Hence
5000 < 3(2^(n- 1))
5000/3 < 2^(n-1)
1666.666... < 2^(n-1)
Take logs
log1666.6666 ... < n - 1log(2)
3.22184... < n(0.30103) - 0.30103
n > = (3.22184 + 0.30103) / 0.30103
n > 10.7027///
nth term is 11.
2018-01-04 12:46 am
3, 6, 12, 24, ...
an = 3 × 2^(n - 1)
a12 = 6144
an = 3 × 2^(n - 1)
a12 = 6144
2018-01-03 9:41 pm
First term a=3
common ratio r= 2
nth term = ar^(n-1)
3 (2)^(n-1) > 5000
2^(n-1) > 5000/3
(n-1) log(2) > log(5000/3)
n log(2) - log(2) > log(5000/3)
n log(2) > log(2) + log(5000/3)
n > ( log(2) + log(5000/3)) / log(2)
n > 11.7
smallest n is 12 such that the nth term is greater than 5000.
common ratio r= 2
nth term = ar^(n-1)
3 (2)^(n-1) > 5000
2^(n-1) > 5000/3
(n-1) log(2) > log(5000/3)
n log(2) - log(2) > log(5000/3)
n log(2) > log(2) + log(5000/3)
n > ( log(2) + log(5000/3)) / log(2)
n > 11.7
smallest n is 12 such that the nth term is greater than 5000.
收錄日期: 2021-04-18 18:00:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180103094343AAqP1aS