正数数列{an}是公差不为零的等差数列,正项数列{bn}是等比数列,a1=b1,a3=b3,a7=b5,若a15=bm,求m的值.

问题描述:

正数数列{an}是公差不为零的等差数列,正项数列{bn}是等比数列,a1=b1,a3=b3,a7=b5,若a15=bm,求m的值.

令an=a1+(n-1)d,bn=b1•qn-1
∵{an}为正数数列
∴d>0
令a1=b1=x
则由a3=b3,a7=b5得:
x+2d=x•q2
x+6d=x•q4
解得
q=

2
,d=
x
2

∴由a15=bm,得
x+14d=x•qm-1
2
m−1
2
=8

m=7.
答案解析:令an=a1+(n-1)d,bn=b1•qn-1,设a1=b1=x,由题意知q=
2
,d=
x
2
,由a15=bm,得x+14d=x•qm-12
m−1
2
=8
,m=7.
考试点:等差数列的性质.
知识点:本题考查数列的性质和应用,解题时要认真审题,仔细解答.