Question

已知{an}是各项为不同的正数的等差数列，lg1，lg2,lg3成等差数列，且a1=2011，求{a1}的通项公式。

Answer 1

∵lga1，lga2，lga3成等差数列
∴lga1+lga3=2lga2=lg(a2)^2
即lg(a1*a3)=lg(a2)^2
∴a1*a3=(a2)^2
而a2=a1+d，a3=a1+2d
∴a1(a1+2d)=(a1+d)^2
∴d^2=0
∴d=0
∴an=a1+(n-1)d=2011

Answer 2

an = a1+(n-1)d
=2011 + (n-1)d
a2 = 2011+d
a3 = 2011 +2d
lg(a3)-lg(a2) = lg(a2) - lg(a1)
lg(a3/a2) = lg(a2/a1)
(2011+2d)/(2011+d) = (2011+d)/2011
(2011+2d)2011 = (2011+d)^2
d=0
an = 2011