Question

数列题，谢谢

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Answer 1

如图

Answer 2

(I)
an =a1.q^(n-1)
|a2-a3|=10
|a1.q -a1.q^2| =10 (1)
a1.a2.a3 = 125
(a1.q) ^3 =125
a1.q = 5 (2)
sub (2) into (1)
|a1.q -a1.q^2| =10
|5 - q| =10
5-q = 10 or -10
q=-5 or 15
from (2)
case 1 : q=-5 , a1=-1
an = -(-5)^(n-1)
case2: q=15 , a1=1/3
an = (1/3).(15)^(n-1)
(II)
case 1: an = -(-5)^(n-1)
bn
= 1/an
= -1/(-5)^(n-1)
Sn
= b1+b2+...+bn
= -[ 1 - (-1/5)^n ] / ( 1 +1/5)
= -(5/6) [ 1 - (-1/5)^n ]
1/a1 +1/a2+...+1/am ≥ 1
-(5/6) [ 1 - (-1/5)^m ] ≥ 1
1 - (-1/5)^m ≤ -6/5
(-1/5)^m ≥ 7/5
m : 无解
case 2: an =(1/3).(15)^(n-1)
bn
= 1/an
= 3. (1/15)^(n-1)
Sn
= b1+b2+...+bn
=3 [ 1 - (1/15)^n ]/ ( 1- 1/15)
=(45/14) [ 1 - (1/15)^n ]
1/a1 +1/a2+...+1/am ≥ 1
(45/14) [ 1 - (1/15)^m ] ≥ 1
1 - (1/15)^m ≥ 14/45
(1/15)^m ≤ 31/45
最小 m =1
=>

最小 m=1
1/a1 +1/a2+...+1/am ≥ 1

追问

第二问哪种情况m都不存在吧。。