Question

已知等比数列{an}中,a1=3,an=48,a(2n-3)=192,则满足

已知等比数列{an}中,a1=3,an=48,a(2n-3)=192,则满足|am|>2009的最小正整数m=?

Answer 1

an=a1*q^(n-1)=48 (i)
a(2n-3)=a1*q^(2n-4)=192 (ii)

(i)=>q^(n-1)=48/3=16 (iii)
(ii)/(i)=> q^(n-3)=192/48=4 (iv)

(iii)/(iv)=>q^2=4
=>|q|=2
|am|>2009
=>|a1*q^(m-1)|>2009
=>|q^(m-1)|>2009/3
又|q|=2
=>|q^9|=512<2009/3
|q^10|=1024>2009/3

因此满足要求的最小m符合：m-1=10
=> m=11

Answer 2

an=3q^(n-1)=48
q^(n-1)=16
a(2n-3)=3q^(2n-4)=192
[q^(n-2)]^2=64
[q^(n-1)/q]^2=64
256/q^2=64
q^2=4
q=2或-2
am=3(±2)^(m-1)
|am|=3*2^(m-1)>2009
2^(m-1)>669
m-1>9
m>10
所以最小正整数m=11

Answer 3

设公比q=2，则An=3*q^n-1=48, A(2n-3)=3*q^(2n-4)=192;
通过上述两式可得n=5，q=2；Am=3*2^M-1 ，令|Am|>2009,m>10,所以M=11