问题描述: 设数列{an}是各项为正等比数列 求证数列{lgan}为等差数列,并写出首项和公差 1个回答 分类:数学 2014-11-16 问题解答: 我来补答 设an=a1*q^n-1则lgan-1+lgan+1=lga1*q^n-2+lga1*qn=lga1^2*q2n-22lgan=2lga1*qn-1=lg(a1*qn-1)^2=lga1^2*q2n-2所以lgan-1+lgan+1=2lgan所以{lgan}是等差数列首项为lga1公差=lgan-lgan-1=lgan/an-1=lgq 展开全文阅读