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
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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
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
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
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
设an公差为d，则 log2Tn=log2（b1b2...bn)=log2b1+log2b2+...+log2bn
=a1+a2+...+an=a1+a1+d+...+a1+(n-1)d=na1+n(n-1)d/2=na2-nd+n(n-1)d/2=log2 1=0.........(1)
由已知可得：a1+a9-2a2=a1+a1+8d-2a1-2d=6d=-28+2*20=12...
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设an公差为d，则 log2Tn=log2（b1b2...bn)=log2b1+log2b2+...+log2bn
=a1+a2+...+an=a1+a1+d+...+a1+(n-1)d=na1+n(n-1)d/2=na2-nd+n(n-1)d/2=log2 1=0.........(1)
由已知可得：a1+a9-2a2=a1+a1+8d-2a1-2d=6d=-28+2*20=12，所以d=2
因此(1)式即：-20n-2n+n(n-1)2/2=-22n+n^2-n=n(n-23)=0，因n>0,所以n-23=0，得n=23
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