已知等差数列{an}中,a2=-20,a1+a9=-28,若数列{bn}满足an=log2bn,设Tn=b1b2...bn,且Tn=1,求n的值

题目

已知等差数列{an}中,a2=-20,a1+a9=-28,若数列{bn}满足an=log2bn,设Tn=b1b2...bn,且Tn=1,求n的值

答案

a2=a1+d=-20
a1+a9=a1+a1+8d=2a1+8d=-28
解方程,得
a1=-22,d=2
an=-22+(n-1)x2=2n-24
an=log2bn=2n-24
bn=2^(2n-24)
Tn=b1b2...bn=2^[2(1+2+3+...+n)-24n]
=2^[2n(n+1)/2-24n]
=2^(n^2-23n)
因为Tn=1,所以n^2-23n=0,故n=0或23
n不能为0,所以n=23