Maths problem: how to do (30), thanks. 我好懷念以前的雅知識,可以留言提問?

首項 a_1 = 2/9,
第4項 a_4 = a_1 r^(4-1) = 6
所以, r^3 = 6/(2/9) = 27
也就是公比 r = 3.

問的是: 
    min(n, a_n > 500}
也就是要找 (2/9)(3^(n-1)) > 500 的最小 n.

3^(n-1) > 500/(2/9) = 2250
3^7 = 2187 < 2250 < 3^8 = 6561
所以, 滿足條件的最小 n 是 9,
a_9 = (2/9)3^8 = 1458.

所以, 大於 500 的最小項是第9項, 其值為 1458.

按: 找 n 一般可以利用對數.
3^(n-1) > 2250
  <==> (n-1) log 3 > log 2250 = log(9000/4)
  <==> (n-1) > log(9000/4)/log(3)
                    = (2×0.477121+3-2×0.301030)/0.477121
                    = 7.03
  <==> n ≧ 9
∴ n 最小值是 9.

ar³ = 6
(2/9) r³ = 6
r³ = 27
r = 3 or -3(rejected)
∴ r = 3

nth term = arⁿ⁻¹ 
　　　　= (2/9)(3)ⁿ⁻¹
　　　　= 2(3)ⁿ⁻³ 

For any term > 500 in the sequence :
2(3)ⁿ⁻³ > 500
(3)ⁿ⁻³ > 250
n-3 > log 250 / log 3
n-3 > 5.03 (to 3 sig.fig)
∴ n > 8.03

So, for the smallest term greater than 500: n=9

∴ The smallest term greater than 500 
= 2(3)⁹⁻³ 
= 2(3)⁶
= 1,458