已知数列{an}的前n项和为Sn,a1=10,a（n+1）=9Sn+10 求证：{lgan}是等差数列

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a(n+1)=9Sn+10
an=9S(n-1)+10
两式相减得
a(n+1)-an=9an
a(n+1)=10an
a(n+1)/an=10
所以an是以10为公比的等比数列
an=a1q^(n-1)
=10*10^(n-1)
=10^n
lgan=lg10^n=n
lga(n-1)=lg10^(n-1)=n-1
lgan-lga(n-1)=1
所以lgan是以1为公差的等差数列

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